A bag contains 4 red balls and 5 green balls.

by

A bag contains 4 red balls and 5 green balls. One ball is drawn at random from the bag. The probability of getting either a red ball or a green ball is
(a) \\ \frac { 4 }{ 9 }
(b) \\ \frac { 5 }{ 9 }
(c) 0
(d) 1

Solution:

In a bag, there are
4 red balls + 5 green balls
Total 4 + 5 = 9
One ball is drawn at random
Probability of either a red or a green ball = \\ \frac { 9 }{ 9 } = 1 (d)

A bag contains 5 red, 4 white and 3 black balls. If a. ball is drawn from the bag at random, then the probability of the ball being not black is
(a) \\ \frac { 5 }{ 12 }
(b) \\ \frac { 1 }{ 3 }
(c) \\ \frac { 3 }{ 4 }
(d) \\ \frac { 1 }{ 4 }

Solution:

In a bag, there are
5 red + 4 white + 3 black balls = 12
One ball is drawn at random
Probability of a ball not black = \\ \frac { 5+4 }{ 12 } = \\ \frac { 9 }{ 12 } = \\ \frac { 3 }{ 4 } (c)

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(a) \\ \frac { 1 }{ 5 }
(b) \\ \frac { 3 }{ 5 }
(c) \\ \frac { 4 }{ 5 }
(d) \\ \frac { 1 }{ 3 }

Solution:

There are t to 40 = 40 tickets in a bag
No. of tickets which is multiple of 5 = 8
(5, 10, 15, 20, 25, 30, 35, 40)
Probability = \\ \frac { 8 }{ 40 } = \\ \frac { 1 }{ 5 } (a)

More Solutions:

Leave a Comment