﻿ Probability Worksheets - Page 5 | Problems & Solutions

Probability Worksheets - Page 5

Probability Worksheets
• Page 5
41.
Find the odds in favor of getting exactly 1 head when three coins are tossed.
 a. 2 : 1 b. 5 : 3 c. 3 : 5 d. 1 : 2

Solution:

Sample space, S = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H), (T, T, T)}

The odds in favor of getting exactly 1 head are 3 : 5
[Three outcomes are successes and 5 are failures.]

42.
A box contains 20 cards labeled 1 through 20. One card is drawn from it at random. Identify the event of drawing a card of an odd number.
 a. {11, 13, 15, 17, 19} b. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} c. {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} d. {1, 3, 5, 7, 9}

Solution:

Each outcome is a number of a card.

Let A be the event of drawing a card of an odd number.

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

The event of drawing a card of an odd number is {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}.

43.
A box contains 20 cards labeled 1 through 20 on them. One card is drawn from it at random. Write the event of drawing a multiple of 5.
 a. {5, 10, 15, 20} b. {5, 10} c. {5, 15} d. {1, 5, 10, 20}

Solution:

Each outcome is a number on a card.

Let B be the event of drawing a card whose number is a multiple of 5.

B = {5, 10, 15, 20}

The event of drawing a card whose number is a multiple of 5 is B = {5, 10, 15, 20}

44.
Fifteen cards labeled A through O are placed in a bag. The bag is shaken and one card is drawn at random. Write the event of drawing a consonant.
 a. {A, B, C, D, E, F, G, H, I, J} b. {B, C, D, F, G, H, J, K, L, M, N} c. {B, C, D, F, G, H, I, J, K, L, M} d. {A, E, I, O}

Solution:

Let Q be the event of drawing a consonant.

Q = {B, C, D, F, G, H, J, K, L, M, N}

45.
What is the probability of getting a red card from a 52-card deck?
 a. $\frac{1}{26}$ b. $\frac{1}{2}$ c. $\frac{1}{13}$ d. $\frac{1}{4}$

Solution:

P(getting a red card) = 26 / 52 = 1 / 2
[There are 26 red cards.]

46.
If two dice are rolled, then what is the probability that the sum of the dots will be 8?
 a. $\frac{5}{36}$ b. $\frac{1}{9}$ c. $\frac{1}{6}$ d. $\frac{5}{6}$

Solution:

The number of elements in the sample space, S = 6 × 6 = 36.

Let E be the event of getting the sum of dots is 8.

E = {(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)}

The number of outcomes in E = 5

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 5 / 36

47.
Four cards are drawn from a bridge deck. What is the probability that all are diamonds?
 a. $\frac{4}{13}$ b. $\frac{11}{4165}$ c. $\frac{1}{4}$ d. $\frac{1}{13}$

Solution:

The sample space, S is the set of all possible 4 card hands in a 52-card deck.

The number of outcomes in S is 52C4 = 52! / 4!48! = 270725

Let E be the event of getting 4 diamonds from 13 in the deck.

The number of outcomes in E is 13C4 = 13! / 4!9! = 715

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 715 / 270725 = 11 / 4165

48.
Two dice are rolled. What is the probability that the sum of the dots showing will be less than 8?
 a. $\frac{1}{2}$ b. $\frac{7}{12}$ c. $\frac{13}{18}$ d. $\frac{5}{9}$

Solution:

The number of elements in sample space is 6 × 6 = 36.

Let E be the event of getting a sum less than 8 on the two dice.

E = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1)}

Number of outcomes in E = 21

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 21 / 36 = 7 / 12

49.
There are 12 boys and 10 girls. If 7 of them are selected at random, what is the probability that 4 are boys and 3 are girls?
 a. $\frac{{}_{7}{C}_{3}}{{}_{12}{C}_{4}}$ b. $\frac{{}_{12}{C}_{7}}{{}_{22}{C}_{7}}$ c. $\frac{\left({}_{12}{C}_{4}\right)\left({}_{10}{C}_{3}\right)}{{}_{22}{C}_{7}}$ d. $\frac{7}{22}$

Solution:

Number of boys = 12

Number of girls = 10

Total number of students = 22

The number of ways of selecting 7 from 22 is 22C7.

So, the number of outcomes in sample space is 22C7.

Let E be the event of selecting 4 boys from 12 and 3 girls from 10.

Number of outcomes in E = 12C4 × 10C3

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= (12C4)(10C3)22C7

50.
Two dice are rolled. What is the probability that both the dice show the same number of dots?
 a. $\frac{1}{6}$ b. 1 c. $\frac{1}{2}$ d. $\frac{1}{3}$

Solution:

The number of elements in sample space = 6 × 6 = 36

Let E be the event of getting the same number of dots on both the dice.

E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Number of outcomes in E = 6

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 6 / 36 = 1 / 6