Question

# There is a box containing two white balls. One more ball was added (either white or...

There is a box containing two white balls. One more ball was added (either white or black, with equal probabilities). Then the balls inside the box were mixed, and one was taken out. It turned out to be white. Given this information, what is the probability that the next ball taken out will also be white?

We are given here that:
P(3 white ) = P(2 white and 1 black) = 0.5

The probability that a white ball is drawn is computed using law of total probability here as:

= 1*0.5 + (2/3)*0.5

= 5/6

Given that the white ball is drawn, the probability that the next ball taken would also be white is computed here as:

We first compute the revised probabilities of having 3 white or 2 white, 1 black using Bayes theorem here as:
P(3 white ) = 0.5*1 / (5/6) = 0.6
P(2 white and 1 black) = 0.5*(2/3) / (5/6) = 0.4

Now the probability that the next ball that will be taken out will also be white is computed here as:

= 1*P(3 white) + (2/3)*(P(2 white and 1 black) )

= 1*0.6 + (2/3)*0.4

= 0.8667

Therefore 0.8667 is the required probability here.

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