Probability — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 13, Probability (সম্ভাৱিতা). It covers Exercise 13.1, Exercise 13.2, Exercise 13.3 and the Miscellaneous Exercise on Chapter 13, working out each conditional probability, independence check and Bayes-theorem problem in full so that you can follow the reasoning and prepare confidently for your examination.
Summary
The conditional probability of an event $E$ given that another event $F$ has already occurred is $P(E \mid F) = \dfrac{P(E \cap F)}{P(F)}$, provided $P(F) \neq 0$. It satisfies $0 \leq P(E \mid F) \leq 1$, $P(E^{\prime} \mid F) = 1 – P(E \mid F)$ and $P((E \cup F) \mid G) = P(E \mid G) + P(F \mid G) – P((E \cap F) \mid G)$. Rearranging the definition gives the multiplication rule $P(E \cap F) = P(F)\,P(E \mid F) = P(E)\,P(F \mid E)$, which extends to three or more events as $P(E \cap F \cap G) = P(E)\,P(F \mid E)\,P(G \mid EF)$.
Two events $E$ and $F$ are independent when the occurrence of one does not change the probability of the other, that is $P(E \cap F) = P(E)\,P(F)$; equivalently $P(E \mid F) = P(E)$ and $P(F \mid E) = P(F)$. Independence is not the same as being mutually exclusive. If $\{E_1, E_2, \dots, E_n\}$ is a partition of the sample space (pairwise disjoint, exhaustive, each of nonzero probability), the theorem of total probability gives $P(A) = \sum_{j=1}^{n} P(E_j)\,P(A \mid E_j)$ for any event $A$.
Bayes theorem reverses the conditioning: for a partition $E_1, E_2, \dots, E_n$ and any event $A$ of nonzero probability, $P(E_i \mid A) = \dfrac{P(E_i)\,P(A \mid E_i)}{\sum_{j=1}^{n} P(E_j)\,P(A \mid E_j)}$. Here the $P(E_i)$ are the priori probabilities of the hypotheses and $P(E_i \mid A)$ is the posteriori probability, so Bayes theorem is the formula for the probability of the causes behind an observed event.
Summary: ASSEB Class 12 Mathematics Chapter 13 Probability explains conditional probability, the multiplication theorem, independent events, the theorem of total probability and Bayes theorem, with complete worked answers to Exercise 13.1, Exercise 13.2, Exercise 13.3 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students preparing for their final examination.
Textbook Questions and Answers
Exercise 13.1
Throughout this exercise we use the definition of conditional probability $P(E \mid F) = \dfrac{P(E \cap F)}{P(F)}$, valid whenever $P(F) \neq 0$.
1. Given that $E$ and $F$ are events such that $P(E) = 0.6$, $P(F) = 0.3$ and $P(E \cap F) = 0.2$, find $P(E \mid F)$ and $P(F \mid E)$.
Answer: $P(E \mid F) = \dfrac{P(E \cap F)}{P(F)} = \dfrac{0.2}{0.3} = \dfrac{2}{3}$ and $P(F \mid E) = \dfrac{P(E \cap F)}{P(E)} = \dfrac{0.2}{0.6} = \dfrac{1}{3}$.
2. Compute $P(A \mid B)$, if $P(B) = 0.5$ and $P(A \cap B) = 0.32$.
Answer: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.32}{0.5} = 0.64$.
3. If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B \mid A) = 0.4$, find (i) $P(A \cap B)$ (ii) $P(A \mid B)$ (iii) $P(A \cup B)$.
Answer: (i) From $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$ we get $P(A \cap B) = P(B \mid A)\,P(A) = 0.4 \times 0.8 = 0.32$.
(ii) $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.32}{0.5} = 0.64$.
(iii) $P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.8 + 0.5 – 0.32 = 0.98$.
4. Evaluate $P(A \cup B)$, if $2P(A) = P(B) = \dfrac{5}{13}$ and $P(A \mid B) = \dfrac{2}{5}$.
Answer: $P(B) = \dfrac{5}{13}$ and $P(A) = \dfrac{1}{2}\,P(B) = \dfrac{5}{26}$. Also $P(A \cap B) = P(A \mid B)\,P(B) = \dfrac{2}{5} \times \dfrac{5}{13} = \dfrac{2}{13}$.
$$P(A \cup B) = P(A) + P(B) – P(A \cap B) = \frac{5}{26} + \frac{5}{13} – \frac{2}{13} = \frac{5}{26} + \frac{10}{26} – \frac{4}{26} = \frac{11}{26}$$
5. If $P(A) = \dfrac{6}{11}$, $P(B) = \dfrac{5}{11}$ and $P(A \cup B) = \dfrac{7}{11}$, find (i) $P(A \cap B)$ (ii) $P(A \mid B)$ (iii) $P(B \mid A)$.
Answer: (i) $P(A \cap B) = P(A) + P(B) – P(A \cup B) = \dfrac{6}{11} + \dfrac{5}{11} – \dfrac{7}{11} = \dfrac{4}{11}$.
(ii) $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{4/11}{5/11} = \dfrac{4}{5}$.
(iii) $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{4/11}{6/11} = \dfrac{4}{6} = \dfrac{2}{3}$.
Determine $P(E \mid F)$ in Exercises 6 to 9.
6. A coin is tossed three times, where (i) $E$: head on third toss, $F$: heads on first two tosses; (ii) $E$: at least two heads, $F$: at most two heads; (iii) $E$: at most two tails, $F$: at least one tail.
Answer: The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$, with $8$ equally likely outcomes.
(i) $E = \{HHH, HTH, THH, TTH\}$, $F = \{HHH, HHT\}$, so $E \cap F = \{HHH\}$. Then $P(F) = \dfrac{2}{8}$, $P(E \cap F) = \dfrac{1}{8}$ and $P(E \mid F) = \dfrac{1/8}{2/8} = \dfrac{1}{2}$.
(ii) $E = \{HHH, HHT, HTH, THH\}$ (at least two heads), $F = \{HHT, HTH, HTT, THH, THT, TTH, TTT\}$ (at most two heads, i.e. all except $HHH$). $E \cap F = \{HHT, HTH, THH\}$. So $P(F) = \dfrac{7}{8}$, $P(E \cap F) = \dfrac{3}{8}$ and $P(E \mid F) = \dfrac{3/8}{7/8} = \dfrac{3}{7}$.
(iii) $E$ = at most two tails = all except $TTT$ = $\{HHH, HHT, HTH, HTT, THH, THT, TTH\}$; $F$ = at least one tail = all except $HHH$ = $\{HHT, HTH, HTT, THH, THT, TTH, TTT\}$. $E \cap F = \{HHT, HTH, HTT, THH, THT, TTH\}$ ($6$ outcomes). So $P(F) = \dfrac{7}{8}$, $P(E \cap F) = \dfrac{6}{8}$ and $P(E \mid F) = \dfrac{6/8}{7/8} = \dfrac{6}{7}$.
7. Two coins are tossed once, where (i) $E$: tail appears on one coin, $F$: one coin shows head; (ii) $E$: no tail appears, $F$: no head appears.
Answer: The sample space is $S = \{HH, HT, TH, TT\}$, four equally likely outcomes.
(i) $E$ = exactly one tail = $\{HT, TH\}$, $F$ = exactly one head = $\{HT, TH\}$, so $E \cap F = \{HT, TH\}$. Then $P(F) = \dfrac{2}{4}$, $P(E \cap F) = \dfrac{2}{4}$ and $P(E \mid F) = \dfrac{2/4}{2/4} = 1$.
(ii) $E$ = no tail = $\{HH\}$, $F$ = no head = $\{TT\}$, so $E \cap F = \varnothing$. Then $P(F) = \dfrac{1}{4}$, $P(E \cap F) = 0$ and $P(E \mid F) = \dfrac{0}{1/4} = 0$.
8. A die is thrown three times, $E$: $4$ appears on the third toss, $F$: $6$ and $5$ appear respectively on first two tosses.
Answer: The experiment has $6^3 = 216$ equally likely outcomes. $F = \{(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)\}$, so $n(F) = 6$. The only outcome of $F$ with $4$ on the third toss is $(6,5,4)$, so $E \cap F = \{(6,5,4)\}$. Hence $P(F) = \dfrac{6}{216}$, $P(E \cap F) = \dfrac{1}{216}$ and $P(E \mid F) = \dfrac{1/216}{6/216} = \dfrac{1}{6}$.
9. Mother, father and son line up at random for a family picture, $E$: son on one end, $F$: father in middle.
Answer: Writing the arrangement as (position 1, position 2, position 3) with $M$ = mother, $Fa$ = father, $So$ = son, the sample space has $3! = 6$ equally likely arrangements. $F$ = father in the middle = $\{(M, Fa, So), (So, Fa, M)\}$, so $n(F) = 2$. In both of these the son occupies an end, so $E \cap F = \{(M, Fa, So), (So, Fa, M)\}$ with $n(E \cap F) = 2$. Hence $P(E \mid F) = \dfrac{2/6}{2/6} = 1$.
10. A black and a red die are rolled. (a) Find the conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$. (b) Find the conditional probability of obtaining the sum $8$, given that the red die resulted in a number less than $4$.
Answer: (a) Let $A$ = sum $> 9$ and $B$ = black die shows $5$. Given the black die is $5$, the possible outcomes are $(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$, so $n(B) = 6$. A sum greater than $9$ needs the red die $> 4$, giving $(5,5)$ and $(5,6)$, so $n(A \cap B) = 2$. Hence $P(A \mid B) = \dfrac{2/36}{6/36} = \dfrac{2}{6} = \dfrac{1}{3}$.
(b) Let $A$ = sum $8$ and $B$ = red die shows a number less than $4$. Writing outcomes as (black, red), $B$ has red $\in \{1, 2, 3\}$, so $n(B) = 6 \times 3 = 18$. Sum $8$ with red $< 4$: $(6,2)$ and $(5,3)$ only, so $n(A \cap B) = 2$. Hence $P(A \mid B) = \dfrac{2/36}{18/36} = \dfrac{2}{18} = \dfrac{1}{9}$.
11. A fair die is rolled. Consider events $E = \{1, 3, 5\}$, $F = \{2, 3\}$ and $G = \{2, 3, 4, 5\}$. Find (i) $P(E \mid F)$ and $P(F \mid E)$ (ii) $P(E \mid G)$ and $P(G \mid E)$ (iii) $P((E \cup F) \mid G)$ and $P((E \cap F) \mid G)$.
Answer: Each outcome has probability $\dfrac{1}{6}$, so $P(E) = \dfrac{3}{6}$, $P(F) = \dfrac{2}{6}$, $P(G) = \dfrac{4}{6}$.
(i) $E \cap F = \{3\}$, so $P(E \cap F) = \dfrac{1}{6}$. Then $P(E \mid F) = \dfrac{1/6}{2/6} = \dfrac{1}{2}$ and $P(F \mid E) = \dfrac{1/6}{3/6} = \dfrac{1}{3}$.
(ii) $E \cap G = \{3, 5\}$, so $P(E \cap G) = \dfrac{2}{6}$. Then $P(E \mid G) = \dfrac{2/6}{4/6} = \dfrac{1}{2}$ and $P(G \mid E) = \dfrac{2/6}{3/6} = \dfrac{2}{3}$.
(iii) $E \cup F = \{1, 2, 3, 5\}$, so $(E \cup F) \cap G = \{2, 3, 5\}$ and $P((E \cup F) \cap G) = \dfrac{3}{6}$; thus $P((E \cup F) \mid G) = \dfrac{3/6}{4/6} = \dfrac{3}{4}$. Also $E \cap F = \{3\}$, so $(E \cap F) \cap G = \{3\}$ and $P((E \cap F) \mid G) = \dfrac{1/6}{4/6} = \dfrac{1}{4}$.
12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?
Answer: Writing (elder, younger), the sample space is $S = \{(b,b), (b,g), (g,b), (g,g)\}$, four equally likely outcomes. Let $E$ = both girls = $\{(g,g)\}$.
(i) $F$ = youngest is a girl = $\{(b,g), (g,g)\}$, so $E \cap F = \{(g,g)\}$. Then $P(E \mid F) = \dfrac{1/4}{2/4} = \dfrac{1}{2}$.
(ii) $F$ = at least one girl = $\{(b,g), (g,b), (g,g)\}$, so $E \cap F = \{(g,g)\}$. Then $P(E \mid F) = \dfrac{1/4}{3/4} = \dfrac{1}{3}$.
13. An instructor has a question bank consisting of $300$ easy True/False questions, $200$ difficult True/False questions, $500$ easy multiple choice questions and $400$ difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer: Let $A$ = easy question and $B$ = multiple choice question. The total number of multiple choice questions is $500 + 400 = 900$, of which $500$ are easy. Since $A \cap B$ = easy multiple choice = $500$ questions out of a bank of $300 + 200 + 500 + 400 = 1400$,
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{500/1400}{900/1400} = \frac{500}{900} = \frac{5}{9}$$
14. Given that the two numbers appearing on throwing two dice are different, find the probability of the event “the sum of numbers on the dice is $4$”.
Answer: Let $A$ = the sum is $4$ and $B$ = the two numbers are different. The two dice give $36$ equally likely outcomes; the $6$ outcomes with equal numbers are excluded, so $n(B) = 30$. Outcomes with sum $4$ are $(1,3), (2,2), (3,1)$; discarding $(2,2)$ (equal numbers) leaves $A \cap B = \{(1,3), (3,1)\}$, so $n(A \cap B) = 2$. Hence $P(A \mid B) = \dfrac{2/36}{30/36} = \dfrac{2}{30} = \dfrac{1}{15}$.
15. Consider the experiment of throwing a die; if a multiple of $3$ comes up, throw the die again, and if any other number comes, toss a coin. Find the conditional probability of the event “the coin shows a tail”, given that “at least one die shows a $3$”.
Answer: Let $E$ = the coin shows a tail and $F$ = at least one die shows a $3$. The event $F$ can only happen when the first throw is a multiple of $3$ (a $3$ itself, or a $6$ followed by a $3$ on the re-throw); in every such case the second action is throwing the die again, so no coin is tossed. Hence a tail on the coin can never occur together with a $3$ on a die, i.e. $E \cap F = \varnothing$ and $P(E \cap F) = 0$. Therefore
$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{P(F)} = 0$$
In each of the Exercises 16 and 17 choose the correct answer.
16. If $P(A) = \dfrac{1}{2}$, $P(B) = 0$, then $P(A \mid B)$ is (A) $0$ (B) $\dfrac{1}{2}$ (C) not defined (D) $1$.
Answer: (C) not defined. Since $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and $P(B) = 0$, the expression involves division by zero and so $P(A \mid B)$ is not defined.
17. If $A$ and $B$ are events such that $P(A \mid B) = P(B \mid A)$, then (A) $A \subset B$ but $A \neq B$ (B) $A = B$ (C) $A \cap B = \varnothing$ (D) $P(A) = P(B)$.
Answer: (D) $P(A) = P(B)$. We have $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$. Setting them equal, and since $P(A \cap B) \neq 0$, we may cancel it to obtain $\dfrac{1}{P(B)} = \dfrac{1}{P(A)}$, i.e. $P(A) = P(B)$.
Exercise 13.2
Two events are independent when $P(A \cap B) = P(A)\,P(B)$; then also $P(A \mid B) = P(A)$ and $P(B \mid A) = P(B)$.
1. If $P(A) = \dfrac{3}{5}$ and $P(B) = \dfrac{1}{5}$, find $P(A \cap B)$ if $A$ and $B$ are independent events.
Answer: For independent events $P(A \cap B) = P(A)\,P(B) = \dfrac{3}{5} \times \dfrac{1}{5} = \dfrac{3}{25}$.
2. Two cards are drawn at random and without replacement from a pack of $52$ playing cards. Find the probability that both the cards are black.
Answer: There are $26$ black cards. Let $E$ = first card black and $F$ = second card black. Then $P(E) = \dfrac{26}{52}$ and, one black card gone, $P(F \mid E) = \dfrac{25}{51}$. By the multiplication rule
$$P(E \cap F) = P(E)\,P(F \mid E) = \frac{26}{52} \times \frac{25}{51} = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$$
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad ones will be approved for sale.
Answer: The box is approved only when all three drawn oranges are good. Drawing without replacement,
$$P(\text{approved}) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{1320}{2730} = \frac{44}{91}$$
4. A fair coin and an unbiased die are tossed. Let $A$ be the event “head appears on the coin” and $B$ be the event “$3$ on the die”. Check whether $A$ and $B$ are independent events or not.
Answer: The sample space has $2 \times 6 = 12$ equally likely outcomes. $A$ (head on the coin) contains $6$ outcomes, so $P(A) = \dfrac{6}{12} = \dfrac{1}{2}$; $B$ ($3$ on the die) contains the $2$ outcomes $(H,3), (T,3)$, so $P(B) = \dfrac{2}{12} = \dfrac{1}{6}$. Now $A \cap B = \{(H,3)\}$, so $P(A \cap B) = \dfrac{1}{12}$. Since $P(A)\,P(B) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12} = P(A \cap B)$, the events $A$ and $B$ are independent.
5. A die marked $1, 2, 3$ in red and $4, 5, 6$ in green is tossed. Let $A$ be the event “the number is even” and $B$ be the event “the number is red”. Are $A$ and $B$ independent?
Answer: $A$ = even = $\{2, 4, 6\}$, so $P(A) = \dfrac{3}{6} = \dfrac{1}{2}$; $B$ = red = $\{1, 2, 3\}$, so $P(B) = \dfrac{3}{6} = \dfrac{1}{2}$. Then $A \cap B = \{2\}$, so $P(A \cap B) = \dfrac{1}{6}$. But $P(A)\,P(B) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} \neq \dfrac{1}{6}$, so $A$ and $B$ are not independent.
6. Let $E$ and $F$ be events with $P(E) = \dfrac{3}{5}$, $P(F) = \dfrac{3}{10}$ and $P(E \cap F) = \dfrac{1}{5}$. Are $E$ and $F$ independent?
Answer: $P(E)\,P(F) = \dfrac{3}{5} \times \dfrac{3}{10} = \dfrac{9}{50}$, while $P(E \cap F) = \dfrac{1}{5} = \dfrac{10}{50}$. Since $\dfrac{9}{50} \neq \dfrac{10}{50}$, we have $P(E \cap F) \neq P(E)\,P(F)$, so $E$ and $F$ are not independent.
7. Given that the events $A$ and $B$ are such that $P(A) = \dfrac{1}{2}$, $P(A \cup B) = \dfrac{3}{5}$ and $P(B) = p$. Find $p$ if they are (i) mutually exclusive (ii) independent.
Answer: (i) If $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$, giving $\dfrac{3}{5} = \dfrac{1}{2} + p$, hence $p = \dfrac{3}{5} – \dfrac{1}{2} = \dfrac{1}{10}$.
(ii) If $A$ and $B$ are independent, $P(A \cap B) = P(A)\,P(B) = \dfrac{p}{2}$, so $P(A \cup B) = P(A) + P(B) – P(A)\,P(B)$ gives $\dfrac{3}{5} = \dfrac{1}{2} + p – \dfrac{p}{2} = \dfrac{1}{2} + \dfrac{p}{2}$. Then $\dfrac{p}{2} = \dfrac{3}{5} – \dfrac{1}{2} = \dfrac{1}{10}$, hence $p = \dfrac{1}{5}$.
8. Let $A$ and $B$ be independent events with $P(A) = 0.3$ and $P(B) = 0.4$. Find (i) $P(A \cap B)$ (ii) $P(A \cup B)$ (iii) $P(A \mid B)$ (iv) $P(B \mid A)$.
Answer: (i) $P(A \cap B) = P(A)\,P(B) = 0.3 \times 0.4 = 0.12$.
(ii) $P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.3 + 0.4 – 0.12 = 0.58$.
(iii) $P(A \mid B) = P(A) = 0.3$ (independence).
(iv) $P(B \mid A) = P(B) = 0.4$ (independence).
9. If $A$ and $B$ are two events such that $P(A) = \dfrac{1}{4}$, $P(B) = \dfrac{1}{2}$ and $P(A \cap B) = \dfrac{1}{8}$, find $P(\text{not } A \text{ and not } B)$.
Answer: $P(\text{not } A \text{ and not } B) = P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 – P(A \cup B)$. Now $P(A \cup B) = \dfrac{1}{4} + \dfrac{1}{2} – \dfrac{1}{8} = \dfrac{2 + 4 – 1}{8} = \dfrac{5}{8}$. Hence the required probability is $1 – \dfrac{5}{8} = \dfrac{3}{8}$.
10. Events $A$ and $B$ are such that $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{7}{12}$ and $P(\text{not } A \text{ or not } B) = \dfrac{1}{4}$. State whether $A$ and $B$ are independent.
Answer: $P(\text{not } A \text{ or not } B) = P(A^{\prime} \cup B^{\prime}) = P((A \cap B)^{\prime}) = 1 – P(A \cap B)$. So $\dfrac{1}{4} = 1 – P(A \cap B)$, giving $P(A \cap B) = \dfrac{3}{4}$. For independence we would need $P(A \cap B) = P(A)\,P(B) = \dfrac{1}{2} \times \dfrac{7}{12} = \dfrac{7}{24}$. Since $\dfrac{3}{4} \neq \dfrac{7}{24}$, the events $A$ and $B$ are not independent.
11. Given two independent events $A$ and $B$ such that $P(A) = 0.3$, $P(B) = 0.6$. Find (i) $P(A \text{ and } B)$ (ii) $P(A \text{ and not } B)$ (iii) $P(A \text{ or } B)$ (iv) $P(\text{neither } A \text{ nor } B)$.
Answer: (i) $P(A \cap B) = P(A)\,P(B) = 0.3 \times 0.6 = 0.18$.
(ii) $A$ and $B^{\prime}$ are also independent, so $P(A \cap B^{\prime}) = P(A)\,P(B^{\prime}) = 0.3 \times 0.4 = 0.12$.
(iii) $P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.3 + 0.6 – 0.18 = 0.72$.
(iv) $P(\text{neither}) = P(A^{\prime} \cap B^{\prime}) = P(A^{\prime})\,P(B^{\prime}) = 0.7 \times 0.4 = 0.28$.
12. A die is tossed thrice. Find the probability of getting an odd number at least once.
Answer: On one toss $P(\text{odd}) = \dfrac{3}{6} = \dfrac{1}{2}$, so $P(\text{even}) = \dfrac{1}{2}$. The three tosses are independent, so $P(\text{no odd in three tosses}) = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$. Hence $P(\text{at least one odd}) = 1 – \dfrac{1}{8} = \dfrac{7}{8}$.
13. Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that (i) both balls are red, (ii) first ball is black and second is red, (iii) one of them is black and other is red.
Answer: With replacement the two draws are independent. From $18$ balls, $P(\text{black}) = \dfrac{10}{18} = \dfrac{5}{9}$ and $P(\text{red}) = \dfrac{8}{18} = \dfrac{4}{9}$.
(i) $P(\text{both red}) = \dfrac{4}{9} \times \dfrac{4}{9} = \dfrac{16}{81}$.
(ii) $P(\text{black then red}) = \dfrac{5}{9} \times \dfrac{4}{9} = \dfrac{20}{81}$.
(iii) $P(\text{one black and one red}) = P(BR) + P(RB) = \dfrac{5}{9} \times \dfrac{4}{9} + \dfrac{4}{9} \times \dfrac{5}{9} = \dfrac{20}{81} + \dfrac{20}{81} = \dfrac{40}{81}$.
14. Probability of solving a specific problem independently by $A$ and $B$ are $\dfrac{1}{2}$ and $\dfrac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.
Answer: Let $P(A) = \dfrac{1}{2}$ and $P(B) = \dfrac{1}{3}$, so $P(A^{\prime}) = \dfrac{1}{2}$ and $P(B^{\prime}) = \dfrac{2}{3}$.
(i) $P(\text{solved}) = P(A \cup B) = 1 – P(A^{\prime})\,P(B^{\prime}) = 1 – \dfrac{1}{2} \times \dfrac{2}{3} = 1 – \dfrac{1}{3} = \dfrac{2}{3}$.
(ii) $P(\text{exactly one}) = P(A)\,P(B^{\prime}) + P(A^{\prime})\,P(B) = \dfrac{1}{2} \times \dfrac{2}{3} + \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{2}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$.
15. One card is drawn at random from a well shuffled deck of $52$ cards. In which of the following cases are the events $E$ and $F$ independent? (i) $E$: the card drawn is a spade, $F$: the card drawn is an ace; (ii) $E$: the card drawn is black, $F$: the card drawn is a king; (iii) $E$: the card drawn is a king or queen, $F$: the card drawn is a queen or jack.
Answer: (i) $P(E) = \dfrac{13}{52} = \dfrac{1}{4}$, $P(F) = \dfrac{4}{52} = \dfrac{1}{13}$. $E \cap F$ = ace of spades = $1$ card, so $P(E \cap F) = \dfrac{1}{52}$. Since $P(E)\,P(F) = \dfrac{1}{4} \times \dfrac{1}{13} = \dfrac{1}{52} = P(E \cap F)$, the events are independent.
(ii) $P(E) = \dfrac{26}{52} = \dfrac{1}{2}$, $P(F) = \dfrac{4}{52} = \dfrac{1}{13}$. $E \cap F$ = black king = $2$ cards, so $P(E \cap F) = \dfrac{2}{52} = \dfrac{1}{26}$. Since $P(E)\,P(F) = \dfrac{1}{2} \times \dfrac{1}{13} = \dfrac{1}{26} = P(E \cap F)$, the events are independent.
(iii) $E$ = king or queen = $8$ cards, $F$ = queen or jack = $8$ cards, so $P(E) = P(F) = \dfrac{8}{52} = \dfrac{2}{13}$. $E \cap F$ = queen = $4$ cards, so $P(E \cap F) = \dfrac{4}{52} = \dfrac{1}{13}$. Now $P(E)\,P(F) = \dfrac{2}{13} \times \dfrac{2}{13} = \dfrac{4}{169}$, while $P(E \cap F) = \dfrac{1}{13} = \dfrac{13}{169}$. Since these differ, the events are not independent.
16. In a hostel, $60\%$ of the students read Hindi newspaper, $40\%$ read English newspaper and $20\%$ read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Answer: Let $H$ = reads Hindi, $E$ = reads English. Then $P(H) = 0.6$, $P(E) = 0.4$, $P(H \cap E) = 0.2$.
(a) $P(\text{neither}) = 1 – P(H \cup E) = 1 – [P(H) + P(E) – P(H \cap E)] = 1 – (0.6 + 0.4 – 0.2) = 1 – 0.8 = 0.2$.
(b) $P(E \mid H) = \dfrac{P(H \cap E)}{P(H)} = \dfrac{0.2}{0.6} = \dfrac{1}{3}$.
(c) $P(H \mid E) = \dfrac{P(H \cap E)}{P(E)} = \dfrac{0.2}{0.4} = \dfrac{1}{2}$.
Choose the correct answer in Exercises 17 and 18.
17. The probability of obtaining an even prime number on each die, when a pair of dice is rolled, is (A) $0$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{12}$ (D) $\dfrac{1}{36}$.
Answer: (D) $\dfrac{1}{36}$. The only even prime number is $2$. The probability of a $2$ on one die is $\dfrac{1}{6}$, and the two dice are independent, so the probability of a $2$ on each is $\dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$.
18. Two events $A$ and $B$ will be independent, if (A) $A$ and $B$ are mutually exclusive (B) $P(A^{\prime}B^{\prime}) = [1 – P(A)][1 – P(B)]$ (C) $P(A) = P(B)$ (D) $P(A) + P(B) = 1$.
Answer: (B) $P(A^{\prime}B^{\prime}) = [1 – P(A)][1 – P(B)]$. If $A$ and $B$ are independent, then $A^{\prime}$ and $B^{\prime}$ are independent, so $P(A^{\prime} \cap B^{\prime}) = P(A^{\prime})\,P(B^{\prime}) = [1 – P(A)][1 – P(B)]$; conversely this condition forces $P(A \cap B) = P(A)\,P(B)$. The other options do not in general imply independence.
Exercise 13.3
Most of these problems use the theorem of total probability $P(A) = \sum_j P(E_j)\,P(A \mid E_j)$ and Bayes theorem $P(E_i \mid A) = \dfrac{P(E_i)\,P(A \mid E_i)}{\sum_j P(E_j)\,P(A \mid E_j)}$.
1. An urn contains $5$ red and $5$ black balls. A ball is drawn at random, its colour is noted and it is returned to the urn. Moreover, $2$ additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Answer: On the first draw $P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2}$ and $P(\text{black}) = \dfrac{1}{2}$. If the first ball is red, the urn becomes $7$ red and $5$ black ($12$ in all), so $P(\text{second red} \mid \text{first red}) = \dfrac{7}{12}$. If the first ball is black, the urn becomes $5$ red and $7$ black, so $P(\text{second red} \mid \text{first black}) = \dfrac{5}{12}$. By total probability,
$$P(\text{second red}) = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2}$$
2. A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Answer: Let $E_1, E_2$ be the events of choosing the first and second bag, so $P(E_1) = P(E_2) = \dfrac{1}{2}$, and let $A$ = drawing a red ball. Then $P(A \mid E_1) = \dfrac{4}{8} = \dfrac{1}{2}$ and $P(A \mid E_2) = \dfrac{2}{8} = \dfrac{1}{4}$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{P(E_1)\,P(A \mid E_1)}{P(E_1)\,P(A \mid E_1) + P(E_2)\,P(A \mid E_2)} = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{4}} = \frac{1/4}{3/8} = \frac{2}{3}$$
3. Of the students in a college, it is known that $60\%$ reside in hostel and $40\%$ are day scholars (not residing in hostel). Previous year results report that $30\%$ of all students who reside in hostel attain A grade and $20\%$ of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the student is a hosteller?
Answer: Let $E_1$ = hosteller, $E_2$ = day scholar, $A$ = A grade. Then $P(E_1) = 0.6$, $P(E_2) = 0.4$, $P(A \mid E_1) = 0.3$, $P(A \mid E_2) = 0.2$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \frac{0.18}{0.18 + 0.08} = \frac{0.18}{0.26} = \frac{9}{13}$$
4. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\dfrac{3}{4}$ be the probability that he knows the answer and $\dfrac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\dfrac{1}{4}$, what is the probability that the student knows the answer given that he answered it correctly?
Answer: Let $E_1$ = knows, $E_2$ = guesses, $A$ = answers correctly. Then $P(E_1) = \dfrac{3}{4}$, $P(E_2) = \dfrac{1}{4}$, $P(A \mid E_1) = 1$ (a student who knows is always correct) and $P(A \mid E_2) = \dfrac{1}{4}$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}} = \frac{3/4}{\frac{3}{4} + \frac{1}{16}} = \frac{3/4}{13/16} = \frac{12}{13}$$
5. A laboratory blood test is $99\%$ effective in detecting a certain disease when it is in fact present. However, the test also yields a false positive result for $0.5\%$ of the healthy persons tested (i.e. if a healthy person is tested, then with probability $0.005$ the test will imply he has the disease). If $0.1\%$ of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Answer: Let $E_1$ = has the disease, $E_2$ = does not have it, $A$ = test is positive. Then $P(E_1) = 0.001$, $P(E_2) = 0.999$, $P(A \mid E_1) = 0.99$, $P(A \mid E_2) = 0.005$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.00099 + 0.004995} = \frac{0.00099}{0.005985} = \frac{22}{133}$$
So the required probability is $\dfrac{22}{133} \approx 0.165$.
6. There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed; it shows heads. What is the probability that it was the two-headed coin?
Answer: Let $E_1, E_2, E_3$ be the events of choosing the two-headed, biased and unbiased coin, each with probability $\dfrac{1}{3}$, and let $A$ = the toss shows heads. Then $P(A \mid E_1) = 1$, $P(A \mid E_2) = \dfrac{3}{4}$, $P(A \mid E_3) = \dfrac{1}{2}$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}} = \frac{1}{1 + \frac{3}{4} + \frac{1}{2}} = \frac{1}{9/4} = \frac{4}{9}$$
7. An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probabilities of an accident are $0.01$, $0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Answer: Of $12000$ drivers, $P(\text{scooter}) = \dfrac{2000}{12000} = \dfrac{1}{6}$, $P(\text{car}) = \dfrac{1}{3}$, $P(\text{truck}) = \dfrac{1}{2}$, with accident probabilities $0.01$, $0.03$, $0.15$. By Bayes theorem,
$$P(\text{scooter} \mid \text{accident}) = \frac{\frac{1}{6} \times 0.01}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.03 + \frac{1}{2} \times 0.15} = \frac{1/600}{1/600 + 6/600 + 45/600} = \frac{1}{52}$$
8. A factory has two machines A and B. Past record shows that machine A produced $60\%$ of the items of output and machine B produced $40\%$ of the items. Further, $2\%$ of the items produced by machine A and $1\%$ produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Answer: Let $E_1$ = produced by A, $E_2$ = produced by B, $D$ = defective. Then $P(E_1) = 0.6$, $P(E_2) = 0.4$, $P(D \mid E_1) = 0.02$, $P(D \mid E_2) = 0.01$. By Bayes theorem,
$$P(E_2 \mid D) = \frac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} = \frac{0.004}{0.012 + 0.004} = \frac{0.004}{0.016} = \frac{1}{4}$$
9. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are $0.6$ and $0.4$ respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$ and the corresponding probability is $0.3$ if the second group wins. Find the probability that the new product introduced was by the second group.
Answer: Let $E_1$ = first group wins, $E_2$ = second group wins, $A$ = a new product is introduced. Then $P(E_1) = 0.6$, $P(E_2) = 0.4$, $P(A \mid E_1) = 0.7$, $P(A \mid E_2) = 0.3$. By Bayes theorem,
$$P(E_2 \mid A) = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54} = \frac{2}{9}$$
10. Suppose a girl throws a die. If she gets a $5$ or $6$, she tosses a coin three times and notes the number of heads. If she gets $1, 2, 3$ or $4$, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw $1, 2, 3$ or $4$ with the die?
Answer: Let $E_1$ = die shows $5$ or $6$ and $E_2$ = die shows $1, 2, 3$ or $4$, so $P(E_1) = \dfrac{2}{6} = \dfrac{1}{3}$ and $P(E_2) = \dfrac{4}{6} = \dfrac{2}{3}$. Let $A$ = exactly one head. For three tosses $P(A \mid E_1) = \binom{3}{1}\left(\dfrac{1}{2}\right)^3 = \dfrac{3}{8}$; for a single toss $P(A \mid E_2) = \dfrac{1}{2}$. By Bayes theorem,
$$P(E_2 \mid A) = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{1/3}{\frac{1}{8} + \frac{1}{3}} = \frac{8/24}{11/24} = \frac{8}{11}$$
11. A manufacturer has three machine operators A, B and C. The first operator A produces $1\%$ defective items, whereas the other two operators B and C produce $5\%$ and $7\%$ defective items respectively. A is on the job for $50\%$ of the time, B is on the job for $30\%$ of the time and C is on the job for $20\%$ of the time. A defective item is produced. What is the probability that it was produced by A?
Answer: Let $E_1, E_2, E_3$ be the events that A, B, C is on the job, with $P(E_1) = 0.5$, $P(E_2) = 0.3$, $P(E_3) = 0.2$, and let $D$ = defective, with $P(D \mid E_1) = 0.01$, $P(D \mid E_2) = 0.05$, $P(D \mid E_3) = 0.07$. By Bayes theorem,
$$P(E_1 \mid D) = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07} = \frac{0.005}{0.005 + 0.015 + 0.014} = \frac{0.005}{0.034} = \frac{5}{34}$$
12. A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer: Let $E_1$ = the lost card is a diamond and $E_2$ = it is not a diamond, so $P(E_1) = \dfrac{13}{52} = \dfrac{1}{4}$ and $P(E_2) = \dfrac{39}{52} = \dfrac{3}{4}$. Let $A$ = the two drawn cards are both diamonds. If the lost card is a diamond, $12$ diamonds remain among $51$ cards, so $P(A \mid E_1) = \dfrac{\binom{12}{2}}{\binom{51}{2}} = \dfrac{66}{1275}$. If the lost card is not a diamond, $13$ diamonds remain, so $P(A \mid E_2) = \dfrac{\binom{13}{2}}{\binom{51}{2}} = \dfrac{78}{1275}$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{\frac{1}{4} \times \frac{66}{1275}}{\frac{1}{4} \times \frac{66}{1275} + \frac{3}{4} \times \frac{78}{1275}} = \frac{66}{66 + 3 \times 78} = \frac{66}{300} = \frac{11}{50}$$
13. Probability that A speaks truth is $\dfrac{4}{5}$. A coin is tossed. A reports that a head appears. The probability that actually there was a head is (A) $\dfrac{4}{5}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{2}{5}$.
Answer: (A) $\dfrac{4}{5}$. Let $E_1$ = a head actually appears and $E_2$ = a tail actually appears, so $P(E_1) = P(E_2) = \dfrac{1}{2}$. Let $R$ = A reports a head. If a head occurred, A reports a head by telling the truth, so $P(R \mid E_1) = \dfrac{4}{5}$; if a tail occurred, A reports a head by lying, so $P(R \mid E_2) = \dfrac{1}{5}$. By Bayes theorem,
$$P(E_1 \mid R) = \frac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2} \times \frac{4}{5} + \frac{1}{2} \times \frac{1}{5}} = \frac{4/10}{4/10 + 1/10} = \frac{4}{5}$$
14. If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$, then which of the following is correct? (A) $P(A \mid B) = \dfrac{P(B)}{P(A)}$ (B) $P(A \mid B) < P(A)$ (C) $P(A \mid B) \geq P(A)$ (D) None of these.
Answer: (C) $P(A \mid B) \geq P(A)$. Since $A \subset B$, we have $A \cap B = A$, so $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{P(A)}{P(B)}$. Because $A \subset B$ gives $P(B) \geq P(A) > 0$, the factor $\dfrac{1}{P(B)} \geq 1$, hence $P(A \mid B) = \dfrac{P(A)}{P(B)} \geq P(A)$.
Miscellaneous Exercise on Chapter 13
1. $A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B \mid A)$, if (i) $A$ is a subset of $B$ (ii) $A \cap B = \varnothing$.
Answer: (i) If $A \subset B$ then $A \cap B = A$, so $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{P(A)}{P(A)} = 1$.
(ii) If $A \cap B = \varnothing$ then $P(A \cap B) = 0$, so $P(B \mid A) = \dfrac{0}{P(A)} = 0$.
2. A couple has two children. (i) Find the probability that both children are males, if it is known that at least one of the children is male. (ii) Find the probability that both children are females, if it is known that the elder child is a female.
Answer: Writing (elder, younger), the sample space is $S = \{(b,b), (b,g), (g,b), (g,g)\}$, four equally likely outcomes.
(i) Let $E$ = both males = $\{(b,b)\}$ and $F$ = at least one male = $\{(b,b), (b,g), (g,b)\}$. Then $E \cap F = \{(b,b)\}$, so $P(E \mid F) = \dfrac{1/4}{3/4} = \dfrac{1}{3}$.
(ii) Let $E$ = both females = $\{(g,g)\}$ and $F$ = elder child is female = $\{(g,b), (g,g)\}$. Then $E \cap F = \{(g,g)\}$, so $P(E \mid F) = \dfrac{1/4}{2/4} = \dfrac{1}{2}$.
3. Suppose that $5\%$ of men and $0.25\%$ of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
Answer: Let $E_1$ = male, $E_2$ = female, so $P(E_1) = P(E_2) = \dfrac{1}{2}$, and let $A$ = grey-haired, with $P(A \mid E_1) = 0.05$ and $P(A \mid E_2) = 0.0025$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{\frac{1}{2} \times 0.05}{\frac{1}{2} \times 0.05 + \frac{1}{2} \times 0.0025} = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525} = \frac{20}{21}$$
4. Suppose that $90\%$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?
Answer: Let $X$ be the number of right-handed people among $10$. This is a binomial variable with $n = 10$, $p = \dfrac{9}{10}$ and $q = \dfrac{1}{10}$, so $P(X = r) = \binom{10}{r}\left(\dfrac{9}{10}\right)^r\left(\dfrac{1}{10}\right)^{10 – r}$. “At most $6$” means $X \leq 6$, and it is easier to use the complement:
$$P(X \leq 6) = 1 – P(X \geq 7) = 1 – \sum_{r=7}^{10} \binom{10}{r}\left(\frac{9}{10}\right)^r\left(\frac{1}{10}\right)^{10 – r}$$
Equivalently, $P(X \leq 6) = \sum_{r=0}^{6} \binom{10}{r}\left(\dfrac{9}{10}\right)^r\left(\dfrac{1}{10}\right)^{10 – r} \approx 0.0128$. (Because $p$ is large, having at most $6$ right-handed people in a sample of $10$ is quite unlikely.)
5. If a leap year is selected at random, what is the chance that it will contain $53$ Tuesdays?
Answer: A leap year has $366$ days $= 52$ weeks and $2$ extra days. These $2$ extra days are one of the $7$ equally likely consecutive pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). The year has $53$ Tuesdays exactly when one of the two extra days is a Tuesday, which happens for (Mon, Tue) and (Tue, Wed) $-$ that is $2$ of the $7$ cases. Hence the probability is $\dfrac{2}{7}$.
6. Suppose we have four boxes A, B, C and D containing coloured marbles as given below. One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B? box C?
| Box | Red | White | Black |
|---|---|---|---|
| A | 1 | 6 | 3 |
| B | 6 | 2 | 2 |
| C | 8 | 1 | 1 |
| D | 0 | 6 | 4 |
Answer: Each box has $10$ marbles and is chosen with probability $\dfrac{1}{4}$. Let $R$ = a red marble is drawn. Then $P(R \mid A) = \dfrac{1}{10}$, $P(R \mid B) = \dfrac{6}{10}$, $P(R \mid C) = \dfrac{8}{10}$, $P(R \mid D) = \dfrac{0}{10} = 0$. The total probability of red is
$$P(R) = \frac{1}{4}\left(\frac{1}{10} + \frac{6}{10} + \frac{8}{10} + 0\right) = \frac{1}{4} \times \frac{15}{10} = \frac{15}{40}$$
By Bayes theorem, $P(A \mid R) = \dfrac{\frac{1}{4} \cdot \frac{1}{10}}{15/40} = \dfrac{1}{15}$, $\;P(B \mid R) = \dfrac{\frac{1}{4} \cdot \frac{6}{10}}{15/40} = \dfrac{6}{15} = \dfrac{2}{5}$, $\;P(C \mid R) = \dfrac{\frac{1}{4} \cdot \frac{8}{10}}{15/40} = \dfrac{8}{15}$.
7. Assume that the chances of a patient having a heart attack is $40\%$. It is also assumed that a meditation and yoga course reduces the risk of heart attack by $30\%$ and prescription of a certain drug reduces its chances by $25\%$. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.
Answer: Let $E_1$ = the patient follows yoga and meditation, $E_2$ = the patient takes the drug, so $P(E_1) = P(E_2) = \dfrac{1}{2}$. Let $A$ = the patient has a heart attack. Yoga reduces the base risk $0.40$ by $30\%$, so $P(A \mid E_1) = 0.40 \times (1 – 0.30) = 0.28$; the drug reduces it by $25\%$, so $P(A \mid E_2) = 0.40 \times (1 – 0.25) = 0.30$. By Bayes theorem,
$$P(E_1 \mid A) = \frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30} = \frac{0.28}{0.28 + 0.30} = \frac{0.28}{0.58} = \frac{14}{29}$$
8. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\dfrac{1}{2}$.)
Answer: A second order determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad – bc$ has four entries $a, b, c, d$, each $0$ or $1$, giving $2^4 = 16$ equally likely determinants. Its value is positive only when $ad – bc > 0$, i.e. $ad = 1$ and $bc = 0$. Now $ad = 1$ requires $a = 1, d = 1$ (one choice), and $bc = 0$ holds for $(b, c) \in \{(0,0), (0,1), (1,0)\}$ (three choices). Hence the number of favourable determinants is $1 \times 3 = 3$, and the required probability is $\dfrac{3}{16}$.
9. An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$, $P(B \text{ fails alone}) = 0.15$, $P(A \text{ and } B \text{ fail}) = 0.15$. Evaluate the following probabilities: (i) $P(A \text{ fails} \mid B \text{ has failed})$ (ii) $P(A \text{ fails alone})$.
Answer: Let $P(A) = P(A \text{ fails}) = 0.2$ and $P(A \cap B) = P(A \text{ and } B \text{ fail}) = 0.15$. Since $B$ fails alone means $B$ fails but not $A$, $P(B \text{ fails alone}) = P(B) – P(A \cap B) = 0.15$, so $P(B) = 0.15 + 0.15 = 0.30$.
(i) $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.15}{0.30} = 0.5$.
(ii) $P(A \text{ fails alone}) = P(A) – P(A \cap B) = 0.20 – 0.15 = 0.05$.
10. Bag I contains $3$ red and $4$ black balls and Bag II contains $4$ red and $5$ black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Answer: Let $E_1$ = a red ball is transferred and $E_2$ = a black ball is transferred, so $P(E_1) = \dfrac{3}{7}$ and $P(E_2) = \dfrac{4}{7}$. After the transfer Bag II has $10$ balls. Let $A$ = the ball drawn from Bag II is red. If a red ball was transferred, Bag II has $5$ red and $5$ black, so $P(A \mid E_1) = \dfrac{5}{10} = \dfrac{1}{2}$; if a black ball was transferred, Bag II has $4$ red and $6$ black, so $P(A \mid E_2) = \dfrac{4}{10} = \dfrac{2}{5}$. By Bayes theorem,
$$P(E_2 \mid A) = \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}} = \frac{8/35}{3/14 + 8/35} = \frac{16/70}{15/70 + 16/70} = \frac{16}{31}$$
Choose the correct answer in each of the following.
11. If $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B \mid A) = 1$, then (A) $A \subset B$ (B) $B \subset A$ (C) $B = \varnothing$ (D) $A = \varnothing$.
Answer: (A) $A \subset B$. From $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)} = 1$ we get $P(A \cap B) = P(A)$, which means $A$ is contained in $B$, i.e. $A \subset B$.
12. If $P(A \mid B) > P(A)$, then which of the following is correct? (A) $P(B \mid A) < P(B)$ (B) $P(A \cap B) < P(A) \cdot P(B)$ (C) $P(B \mid A) > P(B)$ (D) $P(B \mid A) = P(B)$.
Answer: (C) $P(B \mid A) > P(B)$. From $P(A \mid B) > P(A)$, i.e. $\dfrac{P(A \cap B)}{P(B)} > P(A)$, we get $P(A \cap B) > P(A)\,P(B)$. Dividing by $P(A)$ gives $\dfrac{P(A \cap B)}{P(A)} > P(B)$, that is $P(B \mid A) > P(B)$.
13. If $A$ and $B$ are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then (A) $P(B \mid A) = 1$ (B) $P(A \mid B) = 1$ (C) $P(B \mid A) = 0$ (D) $P(A \mid B) = 0$.
Answer: (B) $P(A \mid B) = 1$. The left side is $P(A \cup B)$, so the condition says $P(A \cup B) = P(A)$, which forces $P(B) – P(A \cap B) = 0$, i.e. $P(A \cap B) = P(B)$ (so $B \subset A$). Then $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{P(B)}{P(B)} = 1$.
Additional Questions and Answers
Multiple Choice Questions
1. If $P(A) = 0.4$, $P(B) = 0.5$ and $P(A \cap B) = 0.2$, then $P(A \mid B)$ is (A) $0.4$ (B) $0.5$ (C) $0.2$ (D) $0.8$.
Answer: (A) $0.4$, since $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.5} = 0.4$.
2. If $A$ and $B$ are independent events with $P(A) = 0.3$ and $P(B) = 0.4$, then $P(A \cup B)$ is (A) $0.12$ (B) $0.42$ (C) $0.58$ (D) $0.70$.
Answer: (C) $0.58$. Here $P(A \cap B) = 0.3 \times 0.4 = 0.12$, so $P(A \cup B) = 0.3 + 0.4 – 0.12 = 0.58$.
3. When a die is thrown once, the probability of getting a prime number is (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{2}{3}$.
Answer: (C) $\dfrac{1}{2}$. The prime numbers on a die are $2, 3, 5$, so the probability is $\dfrac{3}{6} = \dfrac{1}{2}$.
4. Two cards are drawn without replacement from a well shuffled pack of $52$ cards. The probability that both are kings is (A) $\dfrac{1}{169}$ (B) $\dfrac{1}{221}$ (C) $\dfrac{4}{663}$ (D) $\dfrac{1}{26}$.
Answer: (B) $\dfrac{1}{221}$, since $P(\text{both kings}) = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{12}{2652} = \dfrac{1}{221}$.
5. Two events with nonzero probabilities that are mutually exclusive are (A) always independent (B) never independent (C) always equally likely (D) always exhaustive.
Answer: (B) never independent. Mutually exclusive events have $P(A \cap B) = 0$, whereas independent events with nonzero probabilities have $P(A \cap B) = P(A)\,P(B) > 0$; the two cannot hold together.
6. For any two events $A$ and $B$ with $P(A) \neq 0$, $P(A \cap B)$ equals (A) $P(A) + P(B)$ (B) $P(A)\,P(B \mid A)$ (C) $P(A) – P(B)$ (D) $P(A \mid B) + P(B)$.
Answer: (B) $P(A)\,P(B \mid A)$. This is the multiplication rule of probability.
7. A bag contains $3$ red and $2$ blue balls. Two balls are drawn at random without replacement. The probability that both are red is (A) $\dfrac{3}{10}$ (B) $\dfrac{9}{25}$ (C) $\dfrac{6}{25}$ (D) $\dfrac{3}{5}$.
Answer: (A) $\dfrac{3}{10}$, since $P(\text{both red}) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = \dfrac{3}{10}$.
8. In Bayes theorem, the conditional probability $P(E_i \mid A)$ is called the (A) priori probability (B) posteriori probability (C) total probability (D) marginal probability.
Answer: (B) posteriori probability. The $P(E_i)$ are the priori probabilities of the hypotheses, and $P(E_i \mid A)$, computed after $A$ is known to have occurred, is the posteriori probability.
Fill in the Blanks
1. If $P(F) \neq 0$, then $P(E \mid F) = $ ________.
Answer: $\dfrac{P(E \cap F)}{P(F)}$.
2. Two events $E$ and $F$ are independent if $P(E \cap F) = $ ________.
Answer: $P(E)\,P(F)$.
3. If $E$ and $F$ are independent events, then $P(E \mid F) = $ ________.
Answer: $P(E)$.
4. $P(E^{\prime} \mid F) = $ ________.
Answer: $1 – P(E \mid F)$.
5. A set of events that are pairwise disjoint, exhaustive and each of nonzero probability is called a ________ of the sample space.
Answer: partition.
True or False
1. If $A$ and $B$ are mutually exclusive events, then $P(A \cap B) = 0$.
Answer: True. Mutually exclusive events cannot occur together, so their intersection is empty.
2. Independent events are always mutually exclusive.
Answer: False. Two independent events with nonzero probabilities have $P(A \cap B) = P(A)\,P(B) > 0$, so they are not mutually exclusive.
3. For any two events, $P(A \mid B)$ is always equal to $P(B \mid A)$.
Answer: False. In general $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ and $P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$ are equal only when $P(A) = P(B)$.
4. If $A \subset B$ with $P(B) \neq 0$, then $P(A \mid B) = \dfrac{P(A)}{P(B)}$.
Answer: True. When $A \subset B$, $A \cap B = A$, so $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{P(A)}{P(B)}$.
5. In Bayes theorem, $P(E_i)$ is called the posteriori probability of the hypothesis $E_i$.
Answer: False. $P(E_i)$ is the priori probability; $P(E_i \mid A)$ is the posteriori probability.
Short Answer Questions
1. A die is thrown twice. Find the probability that the sum of the numbers appearing is $7$.
Answer: The two throws give $36$ equally likely outcomes. Sum $7$ occurs for $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ $-$ that is $6$ outcomes. Hence the probability is $\dfrac{6}{36} = \dfrac{1}{6}$.
2. If $A$ and $B$ are independent events with $P(A) = 0.6$ and $P(B) = 0.4$, find $P(\text{neither } A \text{ nor } B)$.
Answer: $A^{\prime}$ and $B^{\prime}$ are independent, so $P(A^{\prime} \cap B^{\prime}) = P(A^{\prime})\,P(B^{\prime}) = (1 – 0.6)(1 – 0.4) = 0.4 \times 0.6 = 0.24$.
3. A problem is given to three students whose chances of solving it are $\dfrac{1}{2}$, $\dfrac{1}{3}$ and $\dfrac{1}{4}$. Find the probability that the problem is solved.
Answer: Solving independently, $P(\text{none solves}) = \left(1 – \dfrac{1}{2}\right)\left(1 – \dfrac{1}{3}\right)\left(1 – \dfrac{1}{4}\right) = \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} = \dfrac{1}{4}$. Hence $P(\text{solved}) = 1 – \dfrac{1}{4} = \dfrac{3}{4}$.
4. Two balls are drawn at random with replacement from a bag containing $4$ white and $6$ black balls. Find the probability that both balls are white.
Answer: With replacement the draws are independent, and $P(\text{white}) = \dfrac{4}{10} = \dfrac{2}{5}$. Hence $P(\text{both white}) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}$.
Key Terms
| Term | Meaning |
|---|---|
| Conditional probability | The probability of an event $E$ given that another event $F$ has already occurred, $P(E \mid F) = \dfrac{P(E \cap F)}{P(F)}$ with $P(F) \neq 0$. |
| Sample space | The set $S$ of all possible outcomes of a random experiment. |
| Multiplication rule | $P(E \cap F) = P(E)\,P(F \mid E) = P(F)\,P(E \mid F)$, giving the probability that both events occur. |
| Independent events | Events for which $P(E \cap F) = P(E)\,P(F)$; the occurrence of one does not affect the probability of the other. |
| Mutually exclusive events | Events that cannot occur together, so $E \cap F = \varnothing$ and $P(E \cap F) = 0$. |
| Partition of a sample space | A collection $E_1, E_2, \dots, E_n$ of pairwise disjoint, exhaustive events, each with nonzero probability. |
| Theorem of total probability | For a partition $\{E_j\}$ and any event $A$, $P(A) = \sum_{j} P(E_j)\,P(A \mid E_j)$. |
| Bayes theorem | $P(E_i \mid A) = \dfrac{P(E_i)\,P(A \mid E_i)}{\sum_j P(E_j)\,P(A \mid E_j)}$, used to find a reverse (cause) probability. |
| Priori probability | The probability $P(E_i)$ of a hypothesis before the event $A$ is observed. |
| Posteriori probability | The revised probability $P(E_i \mid A)$ of a hypothesis after the event $A$ is known to have occurred. |
| Random variable | A real-valued function whose domain is the sample space of a random experiment. |