Q1. Complete the following statements:

(i) Probability of an event $E$ + Probability of the event “not $E$” = ______.

(ii) The probability of an event that cannot happen is ______. Such an event is called ______.

(iii) The probability of an event that is certain to happen is ______. Such an event is called ______.

(iv) The sum of the probabilities of all the elementary events of an experiment is ______.

(v) The probability of an event is greater than or equal to ______ and less than or equal to ______.

Answer:

(i) $1$   (ii) $0$, impossible event   (iii) $1$, sure (or certain) event   (iv) $1$   (v) $0$, $1$.

Q2. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/He shoots or misses the shot.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.

(iv) A baby is born. It is a boy or a girl.

Answer:

(i) Not equally likely — depends on engine condition, fuel etc.

(ii) Not equally likely — depends on the player’s skill.

(iii) Equally likely — the answer is either right or wrong, with no a-priori bias.

(iv) Equally likely — boy or girl outcomes are taken to be equally likely.

Q3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football match?

Answer: A coin toss is considered fair because the coin is unbiased and has only two possible outcomes — Head or Tail — which are equally likely. Hence each team has the same probability $\dfrac{1}{2}$ of winning the toss, and neither side is favoured.

Q4. Which of the following cannot be the probability of an event? (A) $\dfrac{2}{3}$   (B) $-1.5$   (C) $15\%$   (D) $0.7$

Answer: Probability must satisfy $0 \le P(E) \le 1$. The value $-1.5$ is negative, so it cannot be a probability. Option (B).

Q5. If $P(E) = 0.05$, what is the probability of “not E”?

Answer: $P(\bar{E}) = 1 – P(E) = 1 – 0.05 = 0.95$.

Q6. A bag contains lemon-flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange-flavoured candy? (ii) a lemon-flavoured candy?

Answer:

(i) The bag has no orange candy, so taking out an orange candy is an impossible event. $P = 0$.

(ii) Every candy is lemon flavoured, so taking out a lemon candy is a sure event. $P = 1$.

Q7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is $0.992$. What is the probability that the 2 students have the same birthday?

Answer: Let $E$ = event that 2 students have the same birthday. Then $\bar{E}$ = event that 2 students do not have the same birthday. Given $P(\bar{E}) = 0.992$.

$$P(E) = 1 – P(\bar{E}) = 1 – 0.992 = 0.008$$

Q8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random. What is the probability that the ball drawn is (i) red? (ii) not red?

Answer: Total balls $= 3 + 5 = 8$.

(i) $P(\text{red}) = \dfrac{3}{8}$.

(ii) $P(\text{not red}) = 1 – \dfrac{3}{8} = \dfrac{5}{8}$.

Q9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out at random. What is the probability that it will be (i) red? (ii) white? (iii) not green?

Answer: Total marbles $= 5 + 8 + 4 = 17$.

(i) $P(\text{red}) = \dfrac{5}{17}$.

(ii) $P(\text{white}) = \dfrac{8}{17}$.

(iii) $P(\text{not green}) = 1 – \dfrac{4}{17} = \dfrac{13}{17}$.

Q10. A piggy bank contains hundred 50p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins falls out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? (ii) will not be a Rs 5 coin?

Answer: Total coins $= 100 + 50 + 20 + 10 = 180$.

(i) $P(50p) = \dfrac{100}{180} = \dfrac{5}{9}$.

(ii) Number of Rs 5 coins $= 10$, so $P(\text{Rs 5}) = \dfrac{10}{180} = \dfrac{1}{18}$ and

$$P(\text{not Rs 5}) = 1 – \dfrac{1}{18} = \dfrac{17}{18}$$

Q11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out a fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Answer: Total fish $= 5 + 8 = 13$.

$$P(\text{male fish}) = \dfrac{5}{13}$$

Q12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

Answer: Total outcomes $= 8$.

(i) $P(8) = \dfrac{1}{8}$.

(ii) Odd numbers $= \{1, 3, 5, 7\}$. $P(\text{odd}) = \dfrac{4}{8} = \dfrac{1}{2}$.

(iii) Numbers greater than 2 $= \{3, 4, 5, 6, 7, 8\}$. $P = \dfrac{6}{8} = \dfrac{3}{4}$.

(iv) Numbers less than 9 $= \{1, 2, 3, 4, 5, 6, 7, 8\}$, all 8. $P = \dfrac{8}{8} = 1$ (sure event).

Q13. A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.

Answer: Sample space $= \{1, 2, 3, 4, 5, 6\}$, total $= 6$.

(i) Primes $= \{2, 3, 5\}$, so $P = \dfrac{3}{6} = \dfrac{1}{2}$.

(ii) Numbers strictly between 2 and 6 $= \{3, 4, 5\}$, so $P = \dfrac{3}{6} = \dfrac{1}{2}$.

(iii) Odd numbers $= \{1, 3, 5\}$, so $P = \dfrac{3}{6} = \dfrac{1}{2}$.

Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour, (ii) a face card, (iii) a red face card, (iv) the jack of hearts, (v) a spade, (vi) the queen of diamonds.

Answer: Total cards $= 52$.

(i) Red kings $= 2$. $P = \dfrac{2}{52} = \dfrac{1}{26}$.

(ii) Face cards $= 12$. $P = \dfrac{12}{52} = \dfrac{3}{13}$.

(iii) Red face cards $= 6$. $P = \dfrac{6}{52} = \dfrac{3}{26}$.

(iv) Jack of hearts $= 1$. $P = \dfrac{1}{52}$.

(v) Spades $= 13$. $P = \dfrac{13}{52} = \dfrac{1}{4}$.

(vi) Queen of diamonds $= 1$. $P = \dfrac{1}{52}$.

Q15. Five cards — the ten, jack, queen, king and ace of diamonds — are well-shuffled with their face downwards. One card is then picked up at random. (i) What is the probability that the card is the queen? (ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Answer:

(i) Total cards $= 5$, queens $= 1$. $P(\text{queen}) = \dfrac{1}{5}$.

(ii) Queen is removed; remaining cards $= 4$ (ten, jack, king, ace).

(a) Aces $= 1$. $P(\text{ace}) = \dfrac{1}{4}$.

(b) Queens left $= 0$. $P(\text{queen}) = \dfrac{0}{4} = 0$ (impossible event).

Q16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Answer: Total pens $= 12 + 132 = 144$. Good pens $= 132$.

$$P(\text{good pen}) = \dfrac{132}{144} = \dfrac{11}{12}$$

Q17. (i) A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer:

(i) Total $= 20$, defective $= 4$. $P(\text{defective}) = \dfrac{4}{20} = \dfrac{1}{5}$.

(ii) Now total $= 19$, defective $= 4$, non-defective $= 15$. $P(\text{not defective}) = \dfrac{15}{19}$.

Q18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a perfect square number, (iii) a number divisible by 5.

Answer: Total discs $= 90$.

(i) Two-digit numbers from 10 to 90 $= 81$. $P = \dfrac{81}{90} = \dfrac{9}{10}$.

(ii) Perfect squares from 1 to 90 $= \{1, 4, 9, 16, 25, 36, 49, 64, 81\}$, total $9$. $P = \dfrac{9}{90} = \dfrac{1}{10}$.

(iii) Multiples of 5 from 1 to 90 $= 18$ numbers (5, 10, …, 90). $P = \dfrac{18}{90} = \dfrac{1}{5}$.

Q19. A child has a die whose six faces show the letters A, B, C, D, E, A. The die is thrown once. What is the probability of getting (i) A? (ii) D?

Answer: Total faces $= 6$.

(i) Letter A appears $2$ times. $P(A) = \dfrac{2}{6} = \dfrac{1}{3}$.

(ii) Letter D appears $1$ time. $P(D) = \dfrac{1}{6}$.

Q20. Suppose you drop a die at random on the rectangular region of length 3 m and breadth 2 m. What is the probability that it will land inside a circle of diameter 1 m drawn inside the rectangle?

Answer: Area of rectangle $= 3 \times 2 = 6 \text{ m}^2$. Radius of circle $= 0.5$ m, so area of circle $= \pi (0.5)^2 = \dfrac{\pi}{4} \text{ m}^2$.

$$P(\text{inside circle}) = \dfrac{\pi/4}{6} = \dfrac{\pi}{24}$$

Q21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) she will buy it? (ii) she will not buy it?

Answer: Total $= 144$. Good $= 144 – 20 = 124$.

(i) $P(\text{buy}) = \dfrac{124}{144} = \dfrac{31}{36}$.

(ii) $P(\text{not buy}) = \dfrac{20}{144} = \dfrac{5}{36}$.

Q22. Refer to Example 13. (i) Complete the table for the sums obtained when two dice are thrown together. (ii) A student argues that there are 11 possible outcomes 2, 3, …, 12, hence each has probability $\tfrac{1}{11}$. Do you agree? Justify.

Answer: When two dice are thrown, total outcomes $= 36$. Favourable outcomes for each sum:

(ii) No, the argument is not correct. Although there are 11 possible sums, they are not equally likely. For instance the sum 7 has 6 favourable outcomes while the sum 12 has only 1, so they cannot all have probability $\tfrac{1}{11}$.

Q23. A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Answer: Sample space when a coin is tossed 3 times has $2^3 = 8$ outcomes:

$\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.

Hanif wins on $HHH$ or $TTT$, i.e. 2 outcomes. So,

$$P(\text{win}) = \dfrac{2}{8} = \dfrac{1}{4}, \quad P(\text{lose}) = 1 – \dfrac{1}{4} = \dfrac{3}{4}$$

Q24. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?

Answer: Total outcomes when a die is thrown twice $= 6 \times 6 = 36$.

(i) For 5 not to appear on either throw, each throw has 5 favourable outcomes. So favourable $= 5 \times 5 = 25$.

$$P(\text{no 5}) = \dfrac{25}{36}$$

(ii) $P(\text{5 at least once}) = 1 – P(\text{no 5}) = 1 – \dfrac{25}{36} = \dfrac{11}{36}$.

Q25. Which of the following arguments are correct and which are not? Give reasons. (i) If two coins are tossed simultaneously there are 3 possible outcomes — two heads, two tails or one of each. So each has probability $\tfrac{1}{3}$. (ii) If a die is thrown there are two possible outcomes, an odd number or an even number, so the probability of getting an odd number is $\tfrac{1}{2}$.

Answer:

(i) Incorrect. The actual sample space is $\{HH, HT, TH, TT\}$ with 4 equally likely outcomes. So $P(HH) = \tfrac{1}{4}$, $P(TT) = \tfrac{1}{4}$, $P(\text{one of each}) = \tfrac{2}{4} = \tfrac{1}{2}$. The three “outcomes” stated above are not equally likely.

(ii) Correct. Out of the 6 equally likely outcomes $\{1, 2, 3, 4, 5, 6\}$, three are odd. So $P(\text{odd}) = \tfrac{3}{6} = \tfrac{1}{2}$.

Q1. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?

Answer: Each customer can visit on any of the 5 days. Total ordered outcomes $= 5 \times 5 = 25$.

(i) Both visit on the same day: outcomes $= (T,T), (W,W), (Th,Th), (F,F), (S,S) = 5$.

$$P(\text{same day}) = \dfrac{5}{25} = \dfrac{1}{5}$$

(ii) Consecutive days (in either order): pairs of consecutive days are (T,W), (W,Th), (Th,F), (F,S) — 4 pairs, each can occur in 2 orders, so $4 \times 2 = 8$ outcomes.

$$P(\text{consecutive days}) = \dfrac{8}{25}$$

(iii) Different days $= 1 – P(\text{same day}) = 1 – \dfrac{1}{5} = \dfrac{4}{5}$.

Q2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table that gives a few values of the total score on the two throws. What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6?

Answer: The faces are 1, 2, 2, 3, 3, 6. The completed sum table (each face equally likely) has $6 \times 6 = 36$ entries:

(i) Even totals (2, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 6, 6, 6, 6, 12): counting from the table the even entries number $18$ out of $36$.

$$P(\text{even}) = \dfrac{18}{36} = \dfrac{1}{2}$$

(ii) Total $= 6$ appears $4$ times in the table. $P(6) = \dfrac{4}{36} = \dfrac{1}{9}$.

(iii) Total at least 6: count entries $\ge 6$ — these are 6 (×4), 7 (×2), 8 (×4), 9 (×4), 12 (×1) $= 15$.

$$P(\text{at least 6}) = \dfrac{15}{36} = \dfrac{5}{12}$$

Q3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Answer: Let the number of blue balls be $x$. Total balls $= 5 + x$.

$P(\text{red}) = \dfrac{5}{5+x}$, $P(\text{blue}) = \dfrac{x}{5+x}$.

Given $P(\text{blue}) = 2 \, P(\text{red})$:

$$\dfrac{x}{5+x} = 2 \cdot \dfrac{5}{5+x} \;\Longrightarrow\; x = 10$$

So there are 10 blue balls.

Q4. A box contains 12 balls out of which $x$ are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find $x$.

Answer: $P(\text{black, original}) = \dfrac{x}{12}$.

After adding 6 more black balls, total $= 18$, black $= x + 6$, so $P_2 = \dfrac{x+6}{18}$.

Given $P_2 = 2 P_1$:

$$\dfrac{x+6}{18} = 2 \cdot \dfrac{x}{12} = \dfrac{x}{6}$$

$\Rightarrow 6(x+6) = 18x \Rightarrow 6x + 36 = 18x \Rightarrow 12x = 36 \Rightarrow x = 3$.

So $x = 3$.

Q5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is $\tfrac{2}{3}$. Find the number of blue marbles in the jar.

Answer: Let green marbles $= g$, blue marbles $= 24 – g$.

$P(\text{green}) = \dfrac{g}{24} = \dfrac{2}{3} \Rightarrow g = 16$.

Therefore blue marbles $= 24 – 16 = 8$.

Q. Two coins are tossed simultaneously. Find the probability of getting (i) at least one head, (ii) at most one tail.

Answer: Sample space $= \{HH, HT, TH, TT\}$, total $= 4$.

(i) At least one head $= \{HH, HT, TH\}$, so $P = \dfrac{3}{4}$.

(ii) At most one tail $= \{HH, HT, TH\}$, so $P = \dfrac{3}{4}$.

Q. A card is drawn from a well-shuffled pack of 52 cards. Find the probability that it is (i) neither a king nor a queen, (ii) a black face card.

Answer:

(i) Kings + Queens $= 4 + 4 = 8$. So neither $= 52 – 8 = 44$. $P = \dfrac{44}{52} = \dfrac{11}{13}$.

(ii) Black face cards $=$ J, Q, K of Spades $+$ J, Q, K of Clubs $= 6$. $P = \dfrac{6}{52} = \dfrac{3}{26}$.

Q. A bag contains 6 red, 8 white and 10 black balls. A ball is drawn at random. Find the probability that the ball is (i) white or red, (ii) not black.

Answer: Total $= 24$.

(i) White or red $= 8 + 6 = 14$. $P = \dfrac{14}{24} = \dfrac{7}{12}$.

(ii) Not black $= 24 – 10 = 14$. $P = \dfrac{14}{24} = \dfrac{7}{12}$.

Q. Two dice are rolled simultaneously. Find the probability that the sum on the two dice is a prime number.

Answer: Total outcomes $= 36$. Prime sums possible: $2, 3, 5, 7, 11$.

Counts: 2 → 1, 3 → 2, 5 → 4, 7 → 6, 11 → 2. Total favourable $= 1 + 2 + 4 + 6 + 2 = 15$.

$$P(\text{prime sum}) = \dfrac{15}{36} = \dfrac{5}{12}$$

Q. From the numbers 1 to 25, one number is chosen at random. Find the probability that the chosen number is (i) divisible by 3, (ii) a prime number.

Answer: Total $= 25$.

(i) Multiples of 3 in 1–25: $3, 6, 9, 12, 15, 18, 21, 24 = 8$. $P = \dfrac{8}{25}$.

(ii) Primes in 1–25: $2, 3, 5, 7, 11, 13, 17, 19, 23 = 9$. $P = \dfrac{9}{25}$.

Q. A bag contains tickets numbered 11 to 30. A ticket is taken out at random. Find the probability that the number on the ticket is (i) a multiple of 7, (ii) greater than 25.

Answer: Total tickets $= 20$.

(i) Multiples of 7 in 11 to 30: $14, 21, 28 = 3$. $P = \dfrac{3}{20}$.

(ii) Greater than 25: $26, 27, 28, 29, 30 = 5$. $P = \dfrac{5}{20} = \dfrac{1}{4}$.

Q. The probability that it will rain tomorrow is $0.85$. What is the probability that it will not rain tomorrow?

Answer: $P(\bar{E}) = 1 – 0.85 = 0.15$.

Q. A coin is tossed twice. What is the probability of getting at least one tail?

Answer: Sample space $= \{HH, HT, TH, TT\}$. At least one tail $= \{HT, TH, TT\} = 3$. $P = \dfrac{3}{4}$.

Q. A card is drawn from 52 cards. Find the probability that the card is (i) an ace of red colour, (ii) a black king or a red queen.

Answer:

(i) Red aces $= 2$. $P = \dfrac{2}{52} = \dfrac{1}{26}$.

(ii) Black kings $= 2$, red queens $= 2$, total $= 4$. $P = \dfrac{4}{52} = \dfrac{1}{13}$.

Q. Three coins are tossed together. Find the probability of (i) exactly 2 heads, (ii) at least 2 heads.

Answer: Total outcomes $= 2^3 = 8$.

(i) Exactly 2 heads $= \{HHT, HTH, THH\} = 3$. $P = \dfrac{3}{8}$.

(ii) At least 2 heads $= \{HHT, HTH, THH, HHH\} = 4$. $P = \dfrac{4}{8} = \dfrac{1}{2}$.

Q. A bag contains 5 white, 7 red and 3 black balls. A ball is drawn at random. Find the probability that the ball drawn is (i) white or red, (ii) not black, (iii) neither white nor black.

Answer: Total balls $= 5 + 7 + 3 = 15$.

(i) White or red $= 5 + 7 = 12$. $P = \dfrac{12}{15} = \dfrac{4}{5}$.

(ii) Not black $= 15 – 3 = 12$. $P = \dfrac{12}{15} = \dfrac{4}{5}$.

(iii) Neither white nor black = red only $= 7$. $P = \dfrac{7}{15}$.

Q. A box contains 19 balls bearing numbers 1, 2, 3, …, 19. A ball is drawn at random. Find the probability that the number on the ball is (i) a prime number, (ii) divisible by 3 or 5, (iii) neither divisible by 5 nor by 10, (iv) an even number.

Answer: Total $= 19$.

(i) Primes in 1–19: $\{2, 3, 5, 7, 11, 13, 17, 19\} = 8$. $P = \dfrac{8}{19}$.

(ii) Divisible by 3: $\{3, 6, 9, 12, 15, 18\} = 6$. Divisible by 5: $\{5, 10, 15\} = 3$. Divisible by 15: $\{15\} = 1$. Total = $6 + 3 – 1 = 8$. $P = \dfrac{8}{19}$.

(iii) Multiples of 5: $\{5, 10, 15\}$. (Multiples of 10 are subset of these.) Not divisible by 5 nor 10 $= 19 – 3 = 16$. $P = \dfrac{16}{19}$.

(iv) Even numbers: $\{2, 4, 6, 8, 10, 12, 14, 16, 18\} = 9$. $P = \dfrac{9}{19}$.

Q. From a well-shuffled deck of 52 cards, find the probability of drawing (i) a red king, (ii) a black face card, (iii) a card greater than 6 but less than 10, (iv) a 5 of red colour.

Answer:

(i) Red kings = 2. $P = \dfrac{2}{52} = \dfrac{1}{26}$.

(ii) Black face cards = 6. $P = \dfrac{6}{52} = \dfrac{3}{26}$.

(iii) Cards 7, 8, 9 of all four suits = $3 \times 4 = 12$. $P = \dfrac{12}{52} = \dfrac{3}{13}$.

(iv) Red 5s = 2. $P = \dfrac{2}{52} = \dfrac{1}{26}$.

Q. Two dice are thrown simultaneously. Find the probability of getting (i) the sum as 8, (ii) the sum as a multiple of 4, (iii) at least one 6.

Answer: Total $= 36$.

(i) Sum 8: $\{(2,6), (3,5), (4,4), (5,3), (6,2)\} = 5$. $P = \dfrac{5}{36}$.

(ii) Multiples of 4 (4, 8, 12) — counts: $4 \to 3, 8 \to 5, 12 \to 1$, total $9$. $P = \dfrac{9}{36} = \dfrac{1}{4}$.

(iii) At least one 6: outcomes where first or second is 6. Outcomes with no 6 $= 5 \times 5 = 25$. So with at least one 6 $= 36 – 25 = 11$. $P = \dfrac{11}{36}$.

Q. A bag contains 25 cards numbered 1 to 25. A card is drawn at random. Find the probability that the number on the card is (i) divisible by 3 or 5, (ii) a perfect square, (iii) divisible by 2 and 5 both.

Answer: Total $= 25$.

(i) Multiples of 3 in 1–25: $\{3, 6, 9, 12, 15, 18, 21, 24\} = 8$. Multiples of 5: $\{5, 10, 15, 20, 25\} = 5$. Multiples of 15: $\{15\} = 1$. Total $= 8 + 5 – 1 = 12$. $P = \dfrac{12}{25}$.

(ii) Perfect squares in 1–25: $\{1, 4, 9, 16, 25\} = 5$. $P = \dfrac{5}{25} = \dfrac{1}{5}$.

(iii) Divisible by 2 and 5 both = divisible by 10: $\{10, 20\} = 2$. $P = \dfrac{2}{25}$.

Q. A child’s game has 8 triangles of which 3 are blue and the rest are red, and 10 squares of which 6 are blue and the rest are red. One piece is lost at random. Find the probability that it is a (i) triangle, (ii) square, (iii) red square, (iv) blue triangle.

Answer: Total pieces $= 8 + 10 = 18$.

(i) Triangles $= 8$. $P = \dfrac{8}{18} = \dfrac{4}{9}$.

(ii) Squares $= 10$. $P = \dfrac{10}{18} = \dfrac{5}{9}$.

(iii) Red squares $= 10 – 6 = 4$. $P = \dfrac{4}{18} = \dfrac{2}{9}$.

(iv) Blue triangles $= 3$. $P = \dfrac{3}{18} = \dfrac{1}{6}$.

Q. From a deck of 52 cards, the jacks, queens and kings are removed. From the remaining cards, one card is drawn at random. Find the probability that the card drawn is (i) a face card, (ii) a number card, (iii) an ace, (iv) a black card.

Answer: Cards remaining $= 52 – 12 = 40$.

(i) No face cards remain. $P = 0$.

(ii) Number cards (2, 3, …, 10) of each suit = $9 \times 4 = 36$. Aces are not number cards, so $P(\text{number card}) = \dfrac{36}{40} = \dfrac{9}{10}$.

(iii) Aces = 4. $P = \dfrac{4}{40} = \dfrac{1}{10}$.

(iv) Black cards remaining $= 26 – 6 = 20$. $P = \dfrac{20}{40} = \dfrac{1}{2}$.

Q. A bag contains 5 red balls, 6 white balls and 7 black balls. Two balls are drawn one after another without replacement. Find the probability that one is red and the other is white. (Sequential outcomes counted.)

Answer: Total balls $= 18$. Total ordered draws of 2 balls $= 18 \times 17 = 306$. Favourable: red then white $= 5 \times 6 = 30$; white then red $= 6 \times 5 = 30$. Total favourable $= 60$. $P = \dfrac{60}{306} = \dfrac{10}{51}$.

Q. A box contains 30 cards numbered 1 to 30. One card is drawn at random. Find the probability that the number on the card is (i) divisible by 2 and 3, (ii) divisible by 2 or 3, (iii) a prime number less than 20.

Answer: Total $= 30$.

(i) Divisible by 2 and 3 = divisible by 6: $\{6, 12, 18, 24, 30\} = 5$. $P = \dfrac{5}{30} = \dfrac{1}{6}$.

(ii) Divisible by 2: $15$. Divisible by 3: $10$. Divisible by 6: $5$. Divisible by 2 or 3 $= 15 + 10 – 5 = 20$. $P = \dfrac{20}{30} = \dfrac{2}{3}$.

(iii) Primes less than 20 in 1–30: $\{2, 3, 5, 7, 11, 13, 17, 19\} = 8$. $P = \dfrac{8}{30} = \dfrac{4}{15}$.

Q. The probability of selecting a green marble at random from a jar that contains only green, white and yellow marbles is $\dfrac{1}{4}$. The probability of selecting a white marble at random from the same jar is $\dfrac{1}{3}$. If the jar contains 10 yellow marbles, what is the total number of marbles in the jar?

Answer: Let total marbles $= n$. Then $P(\text{yellow}) = 1 – \dfrac{1}{4} – \dfrac{1}{3} = \dfrac{12 – 3 – 4}{12} = \dfrac{5}{12}$.

$\dfrac{10}{n} = \dfrac{5}{12} \Rightarrow n = 24$. Total marbles $= 24$.

Q. A coin is tossed two times. Find the probability of getting (i) two heads, (ii) at least one head, (iii) no head, (iv) head on the first toss only.

Answer: Sample space $= \{HH, HT, TH, TT\}$, total $4$.

(i) $\{HH\} = 1$. $P = \dfrac{1}{4}$.

(ii) $\{HH, HT, TH\} = 3$. $P = \dfrac{3}{4}$.

(iii) $\{TT\} = 1$. $P = \dfrac{1}{4}$.

(iv) Head on first toss only = $\{HT\} = 1$. $P = \dfrac{1}{4}$.

Q. There are 30 cards of the same size in a bag, numbered from 1 to 30. One card is taken out from the bag at random. Find the probability that the number on the selected card is not divisible by 3.

Answer: Multiples of 3 in 1–30: $\{3, 6, 9, \ldots, 30\}$, count $10$. Not divisible by 3 $= 30 – 10 = 20$. $P = \dfrac{20}{30} = \dfrac{2}{3}$.

Q. From a pack of 52 cards, a card is drawn. Find the probability that the card is (i) a king or a queen, (ii) neither a king nor a queen.

Answer:

(i) Kings + Queens $= 4 + 4 = 8$. $P = \dfrac{8}{52} = \dfrac{2}{13}$.

(ii) Neither $= 52 – 8 = 44$. $P = \dfrac{44}{52} = \dfrac{11}{13}$.

Q. A pair of dice is thrown once. Find the probability that (i) the sum is divisible by 3, (ii) the sum is divisible by 5, (iii) the sum is divisible by 3 and 5.

Answer: Total $= 36$.

(i) Sum divisible by 3 means sum $\in \{3, 6, 9, 12\}$. Counts: $3 \to 2, 6 \to 5, 9 \to 4, 12 \to 1$, total $12$. $P = \dfrac{12}{36} = \dfrac{1}{3}$.

(ii) Sum divisible by 5 means sum $\in \{5, 10\}$. Counts: $5 \to 4, 10 \to 3$, total $7$. $P = \dfrac{7}{36}$.

(iii) Sum divisible by 3 and 5 means divisible by 15. Possible sum $= 15$? Maximum sum is $12$, so impossible. $P = 0$.

Q. A bag contains 18 balls, of which $x$ are white. (i) If a ball is drawn at random, find the probability that it will be a white ball. (ii) If 2 more white balls were placed in the bag, the probability of drawing a white ball would have been twice the original probability. Find $x$.

Answer:

(i) $P_1 = \dfrac{x}{18}$.

(ii) $P_2 = \dfrac{x+2}{20}$. Given $P_2 = 2 P_1$:

$$\dfrac{x+2}{20} = 2 \cdot \dfrac{x}{18} = \dfrac{x}{9}$$

$\Rightarrow 9(x+2) = 20x \Rightarrow 9x + 18 = 20x \Rightarrow 11x = 18 \Rightarrow x = \dfrac{18}{11}$.

Since $x$ must be a whole number, this question has slightly different stated values in some textbooks. With the more standard version where 2 more white balls double the probability, take $x = 3$ when total is 18 (verifying: $\tfrac{3}{18} = \tfrac{1}{6}$ and $\tfrac{5}{20} = \tfrac{1}{4}$ — not exactly double; standard textbook variant gives a clean integer with adjusted numbers). The method shown above is the standard approach.

Q. What is the probability of getting a head when a fair coin is tossed once?

Answer: $P(H) = \dfrac{1}{2}$.

Q. A die is thrown once. What is the probability of getting an odd number?

Answer: Odd numbers $= \{1, 3, 5\}$. $P = \dfrac{3}{6} = \dfrac{1}{2}$.

Q. What is the probability of getting a number greater than 6 when a die is thrown once?

Answer: No outcome greater than 6 exists in $\{1, \ldots, 6\}$. $P = 0$ (impossible event).

Q. What is the probability of getting a number less than 7 when a die is thrown once?

Answer: All 6 outcomes are less than 7. $P = \dfrac{6}{6} = 1$ (sure event).

Q. From a deck of cards, what is the probability of drawing a king?

Answer: $P = \dfrac{4}{52} = \dfrac{1}{13}$.

Q. If $P(E) = \dfrac{2}{5}$, what is $P(\bar{E})$?

Answer: $P(\bar{E}) = 1 – \dfrac{2}{5} = \dfrac{3}{5}$.

Q. A bag contains 4 red and 5 blue balls. Find the probability of drawing a blue ball.

Answer: $P(\text{blue}) = \dfrac{5}{9}$.

Q. What is the probability of getting a tail when a fair coin is tossed?

Answer: $P(T) = \dfrac{1}{2}$.

Q. A card is drawn at random from a pack. What is the probability that it is a heart?

Answer: $P = \dfrac{13}{52} = \dfrac{1}{4}$.

Q. Two coins are tossed. What is the probability of getting two tails?

Answer: $P(TT) = \dfrac{1}{4}$.

Q. A die is thrown. What is the probability of getting a multiple of 2?

Answer: Multiples of 2 in $\{1,\ldots,6\}$ = $\{2, 4, 6\} = 3$. $P = \dfrac{3}{6} = \dfrac{1}{2}$.

Q. A bag has 3 red, 4 blue and 5 green balls. What is the probability of drawing a red ball?

Answer: Total $= 12$. $P(\text{red}) = \dfrac{3}{12} = \dfrac{1}{4}$.

Q. A die is rolled. What is the probability of getting a number which is a factor of 6?

Answer: Factors of 6 in $\{1,\ldots,6\}$ = $\{1, 2, 3, 6\} = 4$. $P = \dfrac{4}{6} = \dfrac{2}{3}$.

Q. A card is drawn from a deck. Find the probability that it is a spade or a heart.

Answer: Spades + Hearts $= 13 + 13 = 26$. $P = \dfrac{26}{52} = \dfrac{1}{2}$.

Q. The probability of an event is $0.62$. What is the probability of its complement?

Answer: $P(\bar{E}) = 1 – 0.62 = 0.38$.

Q. Define the term “sample space” with an example.

Answer: The sample space of a random experiment is the set of all possible outcomes. For example, when a coin is tossed once, the sample space is $S = \{H, T\}$. When a die is rolled, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

Q. What is meant by an “event” in probability theory?

Answer: An event is a subset of the sample space, i.e. a collection of one or more outcomes. For instance, when a die is rolled, the event “getting an even number” corresponds to the subset $\{2, 4, 6\}$ of the sample space.

Q. What are equally likely outcomes? Give an example.

Answer: Two or more outcomes are equally likely when each has the same chance of occurring. For example, the outcomes Head and Tail of a fair coin toss are equally likely. The six outcomes of a fair die are also equally likely.

Q. State the theoretical definition of probability.

Answer: If a random experiment results in $n$ mutually exclusive and equally likely outcomes, of which $m$ are favourable to an event $E$, then the theoretical probability of $E$ is

$$P(E) = \dfrac{m}{n} = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$$

Q. Explain the meaning of “complementary event” with an example.

Answer: If $E$ is an event, then the complementary event $\bar{E}$ is the event “$E$ does not happen”. Together, $E$ and $\bar{E}$ exhaust all possibilities. For example, if $E$ is the event “getting a head” when a coin is tossed, then $\bar{E}$ is “getting a tail”. $P(E) + P(\bar{E}) = 1$.

Q. Distinguish between theoretical and experimental probability.

Answer: Theoretical (classical) probability is calculated using mathematical reasoning and assumes equally likely outcomes: $P(E) = \tfrac{m}{n}$. Experimental (empirical) probability is determined by actually performing the experiment many times and computing $\tfrac{\text{number of times event occurred}}{\text{total number of trials}}$. By the Law of Large Numbers, when the number of trials is large, the experimental probability approaches the theoretical probability.

Q. Why is $P(E) \le 1$ always?

Answer: Since the number of favourable outcomes can never exceed the total number of outcomes, $m \le n$. Therefore $P(E) = \tfrac{m}{n} \le 1$. Equality holds when every outcome is favourable, i.e. when $E$ is the sure event.

Q. State and prove that the sum of probabilities of all elementary events of a random experiment is 1.

Answer: Let the elementary events be $E_1, E_2, \ldots, E_n$, each with one outcome. By definition, $P(E_i) = \tfrac{1}{n}$ for each $i$. Therefore,

$$P(E_1) + P(E_2) + \cdots + P(E_n) = \dfrac{1}{n} + \dfrac{1}{n} + \cdots + \dfrac{1}{n} = \dfrac{n}{n} = 1$$

Q. A bag contains 3 white, 5 red and 7 black balls. A ball is drawn at random. Find the probability that the ball drawn is (i) red, (ii) not white, (iii) neither white nor black.

Answer: Total balls $= 3 + 5 + 7 = 15$.

(i) $P(\text{red}) = \dfrac{5}{15} = \dfrac{1}{3}$.

(ii) $P(\text{not white}) = 1 – \dfrac{3}{15} = \dfrac{12}{15} = \dfrac{4}{5}$.

(iii) Neither white nor black = red only = 5. $P = \dfrac{5}{15} = \dfrac{1}{3}$.

Q. A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card is (i) a heart, (ii) a king of black colour, (iii) not a face card.

Answer:

(i) Hearts $= 13$, $P = \dfrac{13}{52} = \dfrac{1}{4}$.

(ii) Black kings $= 2$, $P = \dfrac{2}{52} = \dfrac{1}{26}$.

(iii) Not face card $= 52 – 12 = 40$, $P = \dfrac{40}{52} = \dfrac{10}{13}$.

Q. Two dice are thrown simultaneously. Find the probability of getting (i) doublets, (ii) sum equal to 7, (iii) sum greater than 9.

Answer: Total outcomes $= 36$.

(i) Doublets $= \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\} = 6$. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

(ii) Sum 7 $= \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} = 6$. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

(iii) Sum greater than 9 means 10, 11, 12. Counts: $10 \to 3, 11 \to 2, 12 \to 1$, total $6$. $P = \dfrac{6}{36} = \dfrac{1}{6}$.

Q. A bag contains 5 red and 8 white balls. A ball is drawn at random. Find the probability that the ball drawn is (i) red, (ii) white, (iii) not red.

Answer: Total $= 13$.

(i) $P(\text{red}) = \dfrac{5}{13}$, (ii) $P(\text{white}) = \dfrac{8}{13}$, (iii) $P(\text{not red}) = \dfrac{8}{13}$.

Q. From a deck of 52 cards, all face cards are removed. Now a card is drawn from the remaining cards. Find the probability that the card drawn is (i) a black card, (ii) a 10, (iii) an ace of clubs.

Answer: Cards remaining $= 52 – 12 = 40$.

(i) Black cards remaining $= 26 – 6 = 20$. $P = \dfrac{20}{40} = \dfrac{1}{2}$.

(ii) Tens $= 4$. $P = \dfrac{4}{40} = \dfrac{1}{10}$.

(iii) Ace of clubs $= 1$. $P = \dfrac{1}{40}$.

Q. A box contains cards numbered from 1 to 100. A card is drawn at random. Find the probability that the number on the card is (i) divisible by 8, (ii) a perfect square, (iii) divisible by 3 or 5.

Answer: Total $= 100$.

(i) Multiples of 8 from 1 to 100: $8, 16, \ldots, 96$ — total $12$. $P = \dfrac{12}{100} = \dfrac{3}{25}$.

(ii) Perfect squares from 1 to 100: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ — total $10$. $P = \dfrac{10}{100} = \dfrac{1}{10}$.

(iii) Multiples of 3: $33$. Multiples of 5: $20$. Multiples of 15: $6$. By inclusion-exclusion: $33 + 20 – 6 = 47$. $P = \dfrac{47}{100}$.

Q. A bag contains 4 red, 6 white and 10 blue balls. A ball is drawn at random. Find the probability that the ball is (i) red or blue, (ii) not red.

Answer: Total $= 20$.

(i) Red or blue $= 4 + 10 = 14$, $P = \dfrac{14}{20} = \dfrac{7}{10}$.

(ii) Not red $= 16$, $P = \dfrac{16}{20} = \dfrac{4}{5}$.

Q. A letter of the English alphabet is chosen at random. Determine the probability that the letter is (i) a vowel, (ii) a consonant.

Answer: Total letters $= 26$. Vowels (a, e, i, o, u) $= 5$.

(i) $P(\text{vowel}) = \dfrac{5}{26}$.

(ii) $P(\text{consonant}) = \dfrac{21}{26}$.

Q. The king, queen and jack of clubs are removed from a deck of 52 cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting (i) a heart, (ii) a queen, (iii) a club.

Answer: Cards remaining $= 52 – 3 = 49$.

(i) Hearts $= 13$, $P = \dfrac{13}{49}$.

(ii) Queens left $= 4 – 1 = 3$. $P = \dfrac{3}{49}$.

(iii) Clubs left $= 13 – 3 = 10$. $P = \dfrac{10}{49}$.

Q. The probability of guessing the correct answer to a certain question is $\dfrac{x}{12}$. If the probability of not guessing the correct answer is $\dfrac{2}{3}$, find $x$.

Answer: $P(\text{correct}) + P(\text{not correct}) = 1$.

$$\dfrac{x}{12} + \dfrac{2}{3} = 1 \Rightarrow \dfrac{x}{12} = \dfrac{1}{3} \Rightarrow x = 4$$

Q. Cards numbered 1, 2, 3, …, 30 are placed in a box. A card is drawn at random. Find the probability that the number on the card is (i) divisible by 2 or 3, (ii) a prime number greater than 7.

Answer: Total $= 30$.

(i) Multiples of 2: $15$. Multiples of 3: $10$. Multiples of 6: $5$. Total $= 15 + 10 – 5 = 20$. $P = \dfrac{20}{30} = \dfrac{2}{3}$.

(ii) Primes greater than 7 in 1–30: $11, 13, 17, 19, 23, 29 = 6$. $P = \dfrac{6}{30} = \dfrac{1}{5}$.

Q. A jar contains 54 marbles each of which is blue, green or white. The probability of selecting a blue marble at random is $\dfrac{1}{3}$ and that of a green marble is $\dfrac{4}{9}$. How many white marbles does the jar contain?

Answer: Blue $= 54 \times \dfrac{1}{3} = 18$. Green $= 54 \times \dfrac{4}{9} = 24$. White $= 54 – (18 + 24) = 12$.

Q. A coin is tossed three times. Find the probability of getting (i) all three heads, (ii) exactly one tail, (iii) at most one head.

Answer: Total outcomes $= 8$.

(i) $\{HHH\}$, $P = \dfrac{1}{8}$.

(ii) Exactly 1 tail $= \{HHT, HTH, THH\} = 3$, $P = \dfrac{3}{8}$.

(iii) At most 1 head $= \{TTT, HTT, THT, TTH\} = 4$, $P = \dfrac{4}{8} = \dfrac{1}{2}$.

Q. A die is rolled twice. Find the probability that the sum of numbers obtained is (i) 9, (ii) 11, (iii) 1.

Answer: Total $= 36$.

(i) Sum 9 $= \{(3,6), (4,5), (5,4), (6,3)\} = 4$, $P = \dfrac{4}{36} = \dfrac{1}{9}$.

(ii) Sum 11 $= \{(5,6), (6,5)\} = 2$, $P = \dfrac{2}{36} = \dfrac{1}{18}$.

(iii) Sum 1 is impossible (smallest possible sum is 2). $P = 0$.

Q. A bag contains $5x$ red and $7$ green balls. If the probability of drawing a red ball is $\dfrac{5}{12}$, find the value of $x$.

Answer: $\dfrac{5x}{5x+7} = \dfrac{5}{12}$.

$60x = 5(5x+7) \Rightarrow 60x = 25x + 35 \Rightarrow 35x = 35 \Rightarrow x = 1$.

Q. The probability of an impossible event is (A) 1   (B) 0   (C) $\tfrac{1}{2}$   (D) $-1$.

Answer: (B) 0. An impossible event has zero favourable outcomes.

Q. The probability of a sure event is (A) 0   (B) $\tfrac{1}{2}$   (C) 1   (D) 2.

Answer: (C) 1. A sure event always occurs.

Q. Which of the following cannot be the probability of an event? (A) $\tfrac{1}{5}$   (B) $0.3$   (C) $\tfrac{7}{4}$   (D) $0$.

Answer: (C) $\tfrac{7}{4}$, since it exceeds 1.

Q. If the probability of an event is $p$, then the probability of its complement is (A) $1 – p$   (B) $p – 1$   (C) $p$   (D) $1/p$.

Answer: (A) $1 – p$.

Q. A die is thrown once. The probability of getting an even prime number is (A) $\tfrac{1}{2}$   (B) $\tfrac{1}{6}$   (C) $\tfrac{1}{3}$   (D) $0$.

Answer: (B) $\tfrac{1}{6}$. The only even prime is 2, occurring once out of 6.

Q. A card is drawn from a deck of 52 cards. The probability of getting a king of red colour is (A) $\tfrac{1}{26}$   (B) $\tfrac{1}{13}$   (C) $\tfrac{1}{4}$   (D) $\tfrac{2}{13}$.

Answer: (A) $\tfrac{1}{26}$. There are 2 red kings out of 52.

Q. The probability of getting exactly two heads in three tosses of a fair coin is (A) $\tfrac{1}{8}$   (B) $\tfrac{3}{8}$   (C) $\tfrac{1}{2}$   (D) $\tfrac{2}{3}$.

Answer: (B) $\tfrac{3}{8}$. Outcomes: HHT, HTH, THH out of 8.

Q. Two dice are thrown simultaneously. The probability of getting a doublet is (A) $\tfrac{1}{6}$   (B) $\tfrac{1}{3}$   (C) $\tfrac{1}{12}$   (D) $\tfrac{5}{36}$.

Answer: (A) $\tfrac{1}{6}$. There are 6 doublets out of 36.

Q. The probability that a leap year has 53 Sundays is (A) $\tfrac{1}{7}$   (B) $\tfrac{2}{7}$   (C) $\tfrac{3}{7}$   (D) $\tfrac{5}{7}$.

Answer: (B) $\tfrac{2}{7}$. A leap year has 366 days = 52 weeks and 2 extra days. The 2 extra days can be (S,M), (M,T), (T,W), (W,Th), (Th,F), (F,Sa), (Sa,Su) — 7 cases. Two of them include a Sunday: (S,M) and (Sa,Su).

Q. A bag contains 3 red and 2 blue marbles. A marble is drawn at random. The probability of drawing a blue marble is (A) $\tfrac{2}{5}$   (B) $\tfrac{1}{4}$   (C) $\tfrac{3}{5}$   (D) $\tfrac{1}{5}$.

Answer: (A) $\tfrac{2}{5}$.

Q. The total number of outcomes when two coins are tossed simultaneously is (A) 2   (B) 3   (C) 4   (D) 8.

Answer: (C) 4. Sample space = $\{HH, HT, TH, TT\}$.

Q. If $P(E) = 0.4$ then $P(\bar{E})$ is (A) 0.4   (B) 0.6   (C) 0.5   (D) 1.

Answer: (B) 0.6.

Q. The probability of getting a face card from a deck of 52 cards is (A) $\tfrac{1}{13}$   (B) $\tfrac{3}{13}$   (C) $\tfrac{1}{4}$   (D) $\tfrac{1}{2}$.

Answer: (B) $\tfrac{3}{13}$. Face cards = 12 out of 52.

Q. The probability of drawing a black queen from a deck of 52 cards is (A) $\tfrac{1}{52}$   (B) $\tfrac{1}{26}$   (C) $\tfrac{1}{13}$   (D) $\tfrac{2}{13}$.

Answer: (B) $\tfrac{1}{26}$. There are 2 black queens out of 52.

Q. A bag has 5 red, 6 white and 7 black balls. The probability of drawing a non-black ball is (A) $\tfrac{7}{18}$   (B) $\tfrac{11}{18}$   (C) $\tfrac{5}{18}$   (D) $\tfrac{6}{18}$.

Answer: Total $= 18$, non-black $= 11$. (B) $\tfrac{11}{18}$.

Q. A fair die is thrown. Find the probability of getting (i) a number divisible by 3, (ii) a number not divisible by 3.

Answer: Total $= 6$. Multiples of 3 in $\{1,\ldots,6\}$ = $\{3, 6\} = 2$.

(i) $P(\text{divisible by 3}) = \dfrac{2}{6} = \dfrac{1}{3}$.

(ii) $P(\text{not divisible by 3}) = 1 – \dfrac{1}{3} = \dfrac{2}{3}$.

Q. Two dice, one red and one green, are rolled. Find the probability that the sum is at least 10.

Answer: Total $= 36$. Sum at least 10 means 10, 11, 12.

Counts: $10 \to 3 \;\{(4,6),(5,5),(6,4)\}$, $11 \to 2 \;\{(5,6),(6,5)\}$, $12 \to 1 \;\{(6,6)\}$. Total $= 6$.

$$P(\text{sum} \ge 10) = \dfrac{6}{36} = \dfrac{1}{6}$$

Q. From a box containing 100 cards numbered 1 to 100, a card is drawn at random. Find the probability that the number on the card is (i) a perfect cube, (ii) a multiple of 11.

Answer: Total $= 100$.

(i) Perfect cubes from 1 to 100: $1, 8, 27, 64 = 4$. $P = \dfrac{4}{100} = \dfrac{1}{25}$.

(ii) Multiples of 11 from 1 to 100: $11, 22, 33, 44, 55, 66, 77, 88, 99 = 9$. $P = \dfrac{9}{100}$.

Q. A bag contains 4 red balls and 3 blue balls. A ball is drawn at random. Find the probability that the ball is (i) red, (ii) not red.

Answer: Total $= 7$.

(i) $P(\text{red}) = \dfrac{4}{7}$.

(ii) $P(\text{not red}) = \dfrac{3}{7}$.

Q. Cards bearing numbers 1, 3, 5, 7, …, 35 are placed in a box. A card is drawn at random. Find the probability that the number on the card is (i) prime, (ii) divisible by 3 and 5 both.

Answer: Cards: $\{1, 3, 5, 7, \ldots, 35\}$ — these are the odd numbers from 1 to 35, total $18$ cards.

(i) Primes among them: $\{3, 5, 7, 11, 13, 17, 19, 23, 29, 31\} = 10$. $P = \dfrac{10}{18} = \dfrac{5}{9}$.

(ii) Divisible by 3 and 5 both = divisible by 15. From the odd cards, $\{15\} = 1$. $P = \dfrac{1}{18}$.

Q. Three coins are tossed simultaneously. Find the probability of getting (i) all heads, (ii) two heads, (iii) at least one head.

Answer: Total $= 8$.

(i) All heads $= \{HHH\} = 1$, $P = \dfrac{1}{8}$.

(ii) Exactly two heads $= \{HHT, HTH, THH\} = 3$, $P = \dfrac{3}{8}$.

(iii) At least one head $= 1 – P(\text{no head}) = 1 – \dfrac{1}{8} = \dfrac{7}{8}$.

Q. A bag contains 4 white, 6 red and 5 black balls. Two balls are drawn at random one by one with replacement. Find the probability that both are red.

Answer: Total $= 15$. With replacement, $P(\text{red, red}) = \dfrac{6}{15} \times \dfrac{6}{15} = \dfrac{36}{225} = \dfrac{4}{25}$.

Q. The probability that A passes an exam is $0.7$ and the probability that B passes the same exam is $0.6$. Assuming the events are independent, what is the probability that both pass?

Answer: $P(A \text{ and } B) = 0.7 \times 0.6 = 0.42$.

Q. A bag contains 6 white, 7 red, and 4 blue balls. A ball is drawn at random. Find the probability that the ball drawn is (i) white, (ii) not red.

Answer: Total $= 17$.

(i) $P(\text{white}) = \dfrac{6}{17}$.

(ii) $P(\text{not red}) = 1 – \dfrac{7}{17} = \dfrac{10}{17}$.

Q. From a deck of 52 cards, find the probability of drawing (i) the king of hearts, (ii) a non-face card.

Answer:

(i) Only one king of hearts. $P = \dfrac{1}{52}$.

(ii) Non-face cards $= 52 – 12 = 40$. $P = \dfrac{40}{52} = \dfrac{10}{13}$.