Probability – Random Balls Questions

Probability solved problems

Problem Statement – A bag contains 30 balls numbered from 1 to 30. One ball is drawn at random. Find the probability that the number of ball is multiple of 5 or 6.

Lets call event the finding a ball which is multiple of 5 as “A”.

Favorable cases = (5,10,15,20,25,30) 

Sample Space = 30 Balls

So Prob(A) = Number of favorable cases / Sample Space = 6/30

Lets call event the finding a ball which is multiple of 6 as “B”.

Number of Favorable cases for event B = (6,12,18,24,30)

Prob (B) = 5/30

Since 30 is multiple of 5 as well 6, therefore events are not mutually exclusive.

P( A and B) = 1/30 (Common multiple = 30)

So, Probability of getting a ball being multiple 5 or 6 is

P(A or B) = P(A) + P(B) – P(AB)

= 6/30 + 5/30 – 1/30 = 10/30 = 1/30

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Problem Statement-2

A bag containing 5 white and 3 black balls. Two balls are drawn at random one after another with replacement. Find the probability that both the balls  drawn are black.

Solution: Probability of drawing black ball in the first draw = P(A) = 3/8

Probability of drawing black ball in the second draw = P(B)= 3/8

Since, the events are independent, the probability that both the balls are black: 

Therefore P(2 Black) = P(1st black) x P(2nd Black)

= 3/8 x 3/8 = 9/64

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Problem Statement 66: A bag contains 5 white and 3 red balls and four balls are successively drawn out and not replaced. What is the chance that (i) white and red balls appear alternatively and (ii) red and white balls appear alternatively?

(i) The probability of drawing a white ball = 5/8

The probability of drawing a red ball = 3/7

The probability of drawing a white ball = 4/6

The probability of drawing a red ball = 2/5

Since, the events are dependent, therefore the required probability is:

P(1W 1R 1W 1R) = 5/8 x 3/7 x 4/6 x 2/5 = 1/14

(ii) The probability of drawing a red ball =  3/8

The probability of drawing a white ball = 5/7

The probability of drawing a red ball = 2/6

The probability of drawing a white ball =4/5

Since, the events are dependent, therefore the required probability is:

P(1W 1R 1W 1R) = 3/8 x 5/7 x 2/6 x 4/5 = 1/14