﻿ Probability Problems - Page 2 | Problems & Solutions

Probability Problems - Page 2

Probability Problems
• Page 2
11.
Find the probability of getting a prize, if in a lottery there are 5 prize and 20 blanks?
 a. $\frac{1}{3}$ b. $\frac{1}{5}$ c. $\frac{1}{4}$ d. $\frac{1}{2}$

Solution:

Given : n(S) = 5
Event of getting prize n(E) = (20+5) = 25

Then,
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{5}{25}$
P(E) = $\frac{1}{5}$

The probability of getting a prize = $\frac{1}{5}$

12.
Find the probability of getting a queen of club or a king of heart. If a card is drawn from ordinary 52 cards.
 a. $\frac{1}{52}$ b. $\frac{1}{13}$ c. $\frac{2}{13}$ d. $\frac{1}{26}$

Solution:

Given : n(S) = 52

Let n(E) = event of getting a queen of club or king of heart = 2
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{2}{52}$
P(E) = $\frac{1}{26}$

The probability of getting a queen of club or a king of heart = $\frac{1}{26}$

13.
Find the probability of getting a double when throwing 2 dice?
 a. $\frac{1}{6}$ b. $\frac{1}{3}$ c. $\frac{1}{5}$ d. $\frac{1}{8}$

Solution:

Given : 2 dice, n(S)= 6$\Times$ 6 = 36
event of getting double, n(E) = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)} =6

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

The probability of getting a queen of club or a king of heart = $\frac{1}{6}$

14.
Three unbiased coins are tossed. Find the probability of getting at least two heads?
 a. $\frac{1}{4}$ b. $\frac{1}{8}$ c. $\frac{1}{2}$ d. $\frac{1}{3}$

Solution:

Let n(S)={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} = 8

Event of getting two heads n(E) = 4

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

The probability of getting at least two heads = $\frac{1}{2}$

15.
Two dice are thrown, the first dice shows 1 and you can not see the other dice. Find the probability of getting 1 on both the dice?
 a. $\frac{1}{36}$ b. $\frac{1}{12}$ c. $\frac{1}{2}$ d. $\frac{1}{6}$

Solution:

Since 1st dice is showing 1
Second n(S)= 6
and event of getting 1, n(E) = 1
so,
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{1)}{6}$

The probability of getting both die as 1 = $\frac{1}{6}$

16.
Find the probability of getting sum of 2 die greater then 8,when first die is 6?
 a. $\frac{1}{2}$ b. $\frac{2}{5}$ c. $\frac{2}{3}$ d. $\frac{1}{3}$

Solution:

Given first die is 6 then,
n(S)={(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}=6
and event of getting greater than 8, n(E) = {(6,3),(6,4),(6,5),(6,6)}=4
so,
P(E) = $\frac{n(4)}{n(6)}$
P(E) = $\frac{2)}{3}$

The probability = $\frac{2}{3}$

17.
Find the probability of getting one head and one tail, when a coin flipped twice?
 a. $\frac{1}{2}$ b. $\frac{1}{3}$ c. $\frac{1}{4}$ d. $\frac{2}{3}$

Solution:

A coin flipped twice so,n(S)=4
and event of getting one head and one tail, n(E) = 2
so,
P(E) = $\frac{n(2)}{n(4)}$
P(E) = $\frac{1)}{2}$

The probability = $\frac{1}{2}$

18.
From a ordinary card pack, 2 cards are drawn together randomly. Find the probability of getting both king cards?
 a. $\frac{1}{225}$ b. $\frac{1}{221}$ c. $\frac{1}{226}$ d. $\frac{1}{223}$

Solution:

As we know ordinary cards = 52

n(S) = $^{52}C_{2}$ =1326
event of getting 2 kings(E) = $^{4}C_{2}$ = 6

P(E) = $\frac{6}{1326}$
P(E) = $\frac{1}{221}$

The probability = $\frac{1}{221}$

19.
Find the probability of getting total sum as prime number, if two dice are tossed?
 a. $\frac{3}{12}$ b. $\frac{4}{12}$ c. $\frac{1}{12}$ d. $\frac{5}{12}$

Solution:

n(S)=6$\Times$ 6=36
event of getting prime n(E) = {(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)} = 15
P(E)=$\frac{15}{36}$
P(E)=$\frac{5}{12}$

The probability = $\frac{5}{12}$

20.
A box 6 jelly 2 red, 1 yellow  and 3 blue. If 1 jelly take out and put it back then take another jelly out. Find the probability of getting 1 blue jelly followed by 1 yellow jelly?
 a. $\frac{1}{6}$ b. $\frac{1}{12}$ c. $\frac{1}{3}$ d. $\frac{1}{9}$

Solution:

Total jelly=6
2 red, 1 yellow  and 3 blue
P(1 blue jelly) = $\frac{3}{6}$=$\frac{1}{2}$
P(1 yellow jelly) = $\frac{1}{6}$
P(E)=$\frac{1}{2}$ $\Times$$\frac{1}{6}$
P(E)=$\frac{1}{12}$

The probability = $\frac{1}{12}$