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  • One urn contains two black balls (labelled B1 and B2) and one white ball. A second urn contains one black ball and two white balls labelled W1 and W2 Suppose the following experiment is performed. One of the two urns is chosen at random.

One urn contains two black balls (labelled B1 and B2) and one white ball. A second urn contains one black ball and two white balls (labelled W1 and W2). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.
(a) Write the sample space showing all possible outcomes
(b) What is the probability that two black balls are chosen?
(c) What is the probability that two balls of opposite colour are chosen?

Answers (1)

Given:

One urn contains 2 black balls & 1 white ball

The other urn contains 1 black ball & 2 white balls

If one of the 2 urns is chosen then a ball is randomly chosen from urn & without replacing the first ball, the second ball is also chosen from the same urn.

  1. Now, we know that,

Sample Space, S = {B1B2, B1W, B2W, B2B1, B2W, WB1, WB2, W1W2, W1B, W2B, W2B, W2W1, BW1, BW2}

Thus, total no. of sample space = 12

  1. Now, if two black balls are chosen,

Favorable outcomes = B1B2 & B2B1

& total no. of favorable outcomes = 2

Now,

Probability = no. of favorable outcomes / total no. of outcomes

                  = 2 /12 = 1 /6

  1. Now, if two balls of opposite colours are chosen,

Favorable outcomes = B1W, B2W, WB1, WB2, W1B, W2B, BW1, BW2

& total no. of favorable outcomes = 8

Now,

Probability = no. of favorable outcomes / total no. of outcomes

                  = 8 /12 = 2 /3

Posted by

infoexpert22

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