﻿ Probability Problems | Problems & Solutions Probability Problems

Probability Problems
• Page 1
1.
Find the probability of getting a sum 7 from 3 throws of a dice? a. $\frac{1}{36}$ b. $\frac{1}{16}$ c. $\frac{1}{9}$ d. $\frac{1}{6}$

Solution:

In 3 throws of a dice n(s) = 6$\times$6$\times$6 = 216
Let E = getting a sum = {(3,4),(5,2),(6,1),(4,3),(2,5),(1,6)}

P(E) = $\frac{n(E)}{n(s)}$
P(E) = $\frac{6}{216}$
P(E) = $\frac{1}{36}$

$\frac{1}{36}$

2.
In a bag you have 11 red,4 blue and 9 orange balls. Suppose you picked up 1 ball randomly. Find the probability that it is neither red nor orange? a. $\frac{1}{12}$ b. $\frac{1}{2}$ c. $\frac{1}{6}$ d. $\frac{1}{3}$

Solution:

Given: total number of balls = 11 + 4 + 9 = 24
n(E) = 4 [blue ball drawn]

P(E) = $\frac{n(E)}{n(s)}$
P(E) = $\frac{4}{24}$
P(E) = $\frac{1}{6}$

Probability that it is neither red nor orange = $\frac{1}{6}$

3.
In a bag there are numbered 1 to 20 tickets then 1 ticket is drawn randomly. Find the probability that ticket drawn has a number which is a multiple of 5 and 7? a. $\frac{1}{3}$ b. $\frac{3}{20}$ c. $\frac{3}{5}$ d. $\frac{3}{10}$

Solution:

Given :  numbers S = {1,2,3,4,.......,18,19,20}
Let E = event of getting a multiple of 5 and 7 = {5,7,10,14,15,20}
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{6}{20}$
P(E) = $\frac{3}{10}$

The probability that ticket drawn has a number which is a multiple of 5 and 7 = $\frac{3}{10}$

4.
Find the probability of getting a face card from an ordinary deck of 52 playing cards? a. $\frac{1}{13}$ b. $\frac{3}{13}$ c. $\frac{4}{13}$ d. $\frac{2}{13}$

Solution:

Given : playing cards n(S) = 52
and n(E)= event of getting a face card = 12
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{12}{52}$
P(E) = $\frac{3}{13}$

The probability of getting a face card = $\frac{3}{13}$

5.
Find the probability of getting a non-face card or a 7 card  from an ordinary deck of 52 playing cards? a. $\frac{10}{13}$ b. $\frac{1}{13}$ c. $\frac{3}{13}$ d. $\frac{7}{13}$

Solution:

Given : playing cards n(S) = 52
and n(non-face)= event of getting a non-face card = 40
n(7)=event of getting a 7 card = 4
n(non-face and 7) = 4/52

P(E) = $\frac{n(E)}{n(S)}$
P(non-face) = $\frac{40}{52}$
P(7) = $\frac{4}{52}$
P(non-face and 7) = $\frac{4}{52}$
P(non-face or 7) = P(non-face)+P(7)-P(non-face and 7)
P(non-face or 7) = $\frac{40}{52}$+ $\frac{4}{52}$- $\frac{4}{52}$
P(non-face or 7) =  $\frac{40}{52}$
P(non-face or 7) =  $\frac{10}{13}$

The probability of getting a non-face card or a 7 card = $\frac{10}{13}$

6.
Tom tossed 2 coins. Find the probability of getting two heads? a. $\frac{1}{3}$ b. $\frac{1}{2}$ c. $\frac{1}{4}$ d. $\frac{1}{6}$

Solution:

As we know each coin has two possible outcomes Head and tail
Here we have 2 coins
So, n(S) = {(H,T)(T,H)(H,H)(T,T)}

P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{1}{4}$

The probability of getting heads = $\frac{1}{4}$

7.
A family has 3 children, if the probability of giving birth to a boy is same as the giving birth to a girl. Find the probability getting at least two boys? a. $\frac{1}{3}$ b. $\frac{1}{2}$ c. $\frac{1}{8}$ d. $\frac{1}{4}$

Solution:

Given : 3 children
Children in order n(S) = {(BBB),(BBG),(BGB),(BGG),(GGG),(GGB),(GBG)(GBB)} = 8
Event of getting 2 boys n(E) = {(BBB),(BBG),(BGB),(GBB)} = 4

P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{4)}{8}$
P(E) = $\frac{1}{2}$

The probability of getting at least 2 boys = P(E) = $\frac{1}{2}$

8.
Find the probability of getting two numbers whose product is odd. If two dice are thrown simultaneously. a. $\frac{2)}{7}$ b. $\frac{1)}{10}$ c. $\frac{1)}{8}$ d. $\frac{2)}{9}$

Solution:

Two dice thrown simultaneously, n(S) = 6$\times$ 6 = 36
Event of getting two numbers product odd n(E) = {(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)} = 8

P(E)=$\frac{n(E)}{n(S)}$
P(E)=$\frac{8}{36}$
P(E)=$\frac{2}{9}$

The probability of getting two numbers whose product is odd = $\frac{2)}{9}$

9.
In a box, there are 7 marbles, 3 black and 4 white. What is the probability that a black marble drawn? a. $\frac{1}{7}$ b. $\frac{2}{7}$ c. $\frac{4}{7}$ d. $\frac{3}{7}$

Solution:

Given marbles n(S) = 7
Black marble n(E) =3

P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{3}{7}$

The probability of getting a black marble = $\frac{3}{7}$

10.
In a room there are 10 girls and 15 boys. Find the probability of selecting at least 2 girls and 1 boy, if 3 children are selected randomly? a. $\frac{23}{90}$ b. $\frac{27}{92}$ c. $\frac{29}{95}$ d. $\frac{24}{93}$

Solution:

Let S = sample space
E = event of selecting 2 girls and 1 boy

n(S) = $^{25}C_{3}$
n(S) = 2300

n1(E) = $^{20}C_{2}$ $\Times$$^{15}C_{1}$
n(E) = 675

Then,
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{675}{2300}$
P(E) = $\frac{27}{92}$

The probability of selecting at least 2 girls and 1 boy = $\frac{27}{92}$