﻿ Probability Worksheets - Page 6 | Problems & Solutions

# Probability Worksheets - Page 6

Probability Worksheets
• Page 6
51.
If the odds in favor of an event are 3 : 5, what is the probability of that event occurring?
 a. $\frac{5}{8}$ b. $\frac{1}{4}$ c. $\frac{3}{8}$ d. $\frac{3}{5}$

#### Solution:

The odds in favor of an event are = 3 : 5
[number of successes : number of failures]

Therefore, the probability of an event to occur = 3 / 8

52.
A bag consists of 5 red, 4 white balls. If 3 balls are drawn at random, what is the probability that they are red?
 a. $\frac{1}{3}$ b. $\frac{5}{42}$ c. $\frac{3}{5}$ d. $\frac{5}{9}$

#### Solution:

Total number of balls = 9
[5 red, 4 white.]

The number of ways of drawing 3 balls from 9 is 9C3

So, the number of outcomes in sample space = 9C3 = 9! / 3!6! = 84

Let E be the event of drawing 3 red balls from 5.

Number of outcomes in E = 5C3 = 10

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

= 10 / 84 = 5 / 42

53.
Two dice are rolled. What is the probability of that both show a prime number of dots?
 a. $\frac{1}{4}$ b. $\frac{1}{6}$ c. $\frac{1}{2}$ d. $\frac{1}{12}$

#### Solution:

When two dice are rolled, the sample space consists of 6 × 6 = 36 elements.

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

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

Number of outcomes in E = 9

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

= 9 / 36 = 1 / 4

54.
A 3-digit number is formed with the digits 1, 2, 3, 4, 5, 6 with no repetition of any digit. Find the probability that the number is divisible by 5?
 a. 1 b. $\frac{5}{6}$ c. $\frac{1}{2}$ d. $\frac{1}{6}$

#### Solution:

The three digit numbers formed with the digits 1, 2, 3, 4, 5, 6 with no repetition of any digit in 6P3 ways.

= 6! / 3! = 120 ways

Let E be the event that the 3-digit number is divisible by 5.

Number of outcomes in E = 5P2 = 20

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

= 20 / 120 = 1 / 6

55.
The letters of the word PLANE are arranged at random. Find the probability of the words formed to start with A and end with P.
 a. $\frac{1}{20}$ b. $\frac{1}{5}$ c. $\frac{1}{2}$ d. $\frac{3}{5}$

#### Solution:

The letters of the word PLANE can be arranged in 5P5 ways.

= 5! ways.

Number of outcomes in sample space is 5! = 120.

Let E be the event of words formed by the letters at PLANE so that they start with A and ends with P.

Number of outcomes in E = 3! = 6

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

= 6 / 120 = 1 / 20

56.
Sam, Andy, and Clara are playing a game. Each has to roll a die thrice. In each roll, if the die stops on red portion, then Sam will win. If it stops on blue, then Andy wins and if it stops on green area, then Clara would win the game. Find the probability of Andy winning the game.

 a. 1 b. $\frac{1}{27}$ c. $\frac{1}{3}$ d. $\frac{1}{9}$

57.
What is the probability of an arrow shot landing in the white portion?

 a. $\frac{1}{2}$ b. 75% c. 25% d. 0.75%

58.
Matt dropped a counter at random. What is the probability that it will fall in the shaded region?

 a. 36% b. $\frac{1}{25}$ c. 25% d. $\frac{45}{70}$

59.
The circle shown is of radius 3 units. What is the probability of a dart thrown landing inside the circle but outside the triangle?

 a. about 1.71 b. about 59% c. about 0.59% d. about $\frac{1}{2}$

 a. $\frac{1}{9}$ b. $\frac{9}{16}$ c. $\frac{1}{25}$ d. $\frac{9}{25}$