Question

48 40) An urn contains 5 red, 4 white and 3 black balls. The probability of three balls being of different colors when the ball is replaced after each draw is equal to: Independent a) 3/144 b) 4/144 c) 5/144 d) 1​

Answer 1

Given :  An urn contains 5 red, 4 white and 3 black balls.

To Find :  The probability of three balls being of different colors when the ball is replaced after each draw is equal to

Solution:

R = 5

W = 4

B = 3

Balls are replaced after each draw Hence probability of drawing a ball does not change

P(R) = 5/12

P(W) = 4/12

P( B) = 3/12

Probability of Each ball of Different color = (5/12)(4/12)(3/12)

= 5/144

The probability of three balls being of different colors = 5/144

correct option is c) 5/144

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Answer 2

Given,

An urn with balls

• 5 red balls

• 4 white balls

• 3 black balls

To find,  

Probability of three balls being of different colors when the ball is replaced after each draw.  

Solution,  

Total number of balls in the urn = 5 + 4 + 3 = 12 balls  

Total cases possible when three balls are drawn after replacement  

Now, for all three balls to be distinct, each of the three balls has to be chosen in any possible order.

Choices available for red ball = 5  

Choices available for white ball = 4

Choices available for black ball = 3  

Total possible cases satisfying the condition = 5*4*3 = 60  

P(three balls being of different colors when the ball is replaced after each draw) = $\frac{60}{1728} = \frac{5}{144}$  

Therefore, the probability for the given conditions is = $\frac{5}{144}$