Probability Questions and Answers

Q: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

A product is odd only when both numbers are odd, so use the complement.\[ P(\text{both odd}) = \frac{3}{6} \times \frac{3}{6} = \frac{1}{4} \]\[ P(\text{product even}) = 1 - \frac{1}{4} = \frac{3}{4} \]A single even factor is enough to make the product even, which is why the complement is so much easier to count.

Q: In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

There are \( 10 + 25 = 35 \) tickets in the lottery.\[ P = \frac{10}{35} = \frac{2}{7} \]The denominator is the total number of tickets, not the number of blanks - using 25 would give \( \tfrac{2}{5} \).

Q: From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

There are four kings in the pack.\[ P = \frac{{}^{4}C_{2}}{{}^{52}C_{2}} = \frac{6}{1326} = \frac{1}{221} \]Equivalently \( \tfrac{4}{52} \times \tfrac{3}{51} = \tfrac{1}{221} \), the second fraction reflecting that the first card is not replaced.

Q: A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

The bag holds \( 4 + 5 + 6 = 15 \) balls, 5 of them red.\[ P = \frac{{}^{5}C_{3}}{{}^{15}C_{3}} = \frac{10}{455} = \frac{2}{91} \]

Q: Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

Choose one of the 13 spades and one of the 13 hearts.\[ P = \frac{13 \times 13}{{}^{52}C_{2}} = \frac{169}{1326} = \frac{13}{102} \]The two cards are drawn together, so the pair is unordered and the denominator uses a combination; the numerator needs no factor of 2, since choosing one from each suit already counts each pair once.

Q: A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

The bag holds \( 6 + 8 = 14 \) balls.\[ P = \frac{8}{14} = \frac{4}{7} \]The denominator is the total, not the number of black balls - using 6 would give \( \tfrac{4}{3} \), which is not even a valid probability.

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111.

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

A.1/2
B.3/4
C.3/8
D.5/16

Answer: B

A product is odd only when both numbers are odd, so use the complement.

\[ P(\text{both odd}) = \frac{3}{6} \times \frac{3}{6} = \frac{1}{4} \]

\[ P(\text{product even}) = 1 - \frac{1}{4} = \frac{3}{4} \]

A single even factor is enough to make the product even, which is why the complement is so much easier to count.

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112.

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A.21/46
B.25/117
C.1/50
D.3/25

Answer: A

Choose one girl from 10 and two boys from 15, out of all ways to choose 3 from 25.

\[ P = \frac{{}^{10}C_{1} \times {}^{15}C_{2}}{{}^{25}C_{3}} = \frac{10 \times 105}{2300} = \frac{21}{46} \]

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113.

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A.1/10
B.2/5
C.2/7
D.5/7

Answer: C

There are \( 10 + 25 = 35 \) tickets in the lottery.

\[ P = \frac{10}{35} = \frac{2}{7} \]

The denominator is the total number of tickets, not the number of blanks - using 25 would give \( \tfrac{2}{5} \).

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114.

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

A.1/15
B.25/57
C.35/256
D.1/221

Answer: D

There are four kings in the pack.

\[ P = \frac{{}^{4}C_{2}}{{}^{52}C_{2}} = \frac{6}{1326} = \frac{1}{221} \]

Equivalently \( \tfrac{4}{52} \times \tfrac{3}{51} = \tfrac{1}{221} \), the second fraction reflecting that the first card is not replaced.

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115.

Two dice are tossed. The probability that the total score is a prime number is:

A.1/6
B.5/12
C.1/2
D.7/9

Answer: B

Two dice give 36 outcomes. The prime totals are 2, 3, 5, 7 and 11:

\[ 1 + 2 + 4 + 6 + 2 = 15 \text{ outcomes} \]

\[ P = \frac{15}{36} = \frac{5}{12} \]

9 is not prime, and neither is 1 - the two totals most often included by mistake.

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116.

A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

A.1/13
B.2/13
C.1/26
D.1/52

Answer: C

There is exactly one queen of clubs and one king of hearts, and no card is both.

\[ P = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26} \]

The question names specific cards, not whole ranks, so the count is 2 rather than 8.

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117.

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A.1/22
B.3/22
C.2/91
D.2/77

Answer: C

The bag holds \( 4 + 5 + 6 = 15 \) balls, 5 of them red.

\[ P = \frac{{}^{5}C_{3}}{{}^{15}C_{3}} = \frac{10}{455} = \frac{2}{91} \]

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118.

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

A.3/20
B.29/34
C.47/100
D.13/102

Answer: D

Choose one of the 13 spades and one of the 13 hearts.

\[ P = \frac{13 \times 13}{{}^{52}C_{2}} = \frac{169}{1326} = \frac{13}{102} \]

The two cards are drawn together, so the pair is unordered and the denominator uses a combination; the numerator needs no factor of 2, since choosing one from each suit already counts each pair once.

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119.

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

A.1/13
B.3/13
C.1/4
D.9/52

Answer: B

Each suit has three face cards - jack, queen and king - so there are \( 3 \times 4 = 12 \) in the pack.

\[ P = \frac{12}{52} = \frac{3}{13} \]

The ace is not a face card here, as the question makes explicit; counting it would give \( \tfrac{16}{52} \).

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120.

A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A.3/4
B.4/7
C.1/8
D.3/7

Answer: B

The bag holds \( 6 + 8 = 14 \) balls.

\[ P = \frac{8}{14} = \frac{4}{7} \]

The denominator is the total, not the number of black balls - using 6 would give \( \tfrac{4}{3} \), which is not even a valid probability.

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