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Conditional Probability & Bayes' Theorem · Question 99 of 125

Question 99

HardOriginal question

Bag I has 4 white and 2 black balls; bag II has 2 white and 4 black balls. Two balls are drawn at random from bag I and put into bag II. A ball is then drawn at random from bag II. What is the probability that it is white?

Options for question 99

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. The two balls moved are both white, one of each, or both black, with probabilities 615\frac{6}{15}, 815\frac{8}{15} and 115\frac{1}{15}; bag II then holds 4, 3 or 2 white balls of 8.

  2. 615×48+815×38+115×28=24+24+2120=512\frac{6}{15} \times \frac{4}{8} + \frac{8}{15} \times \frac{3}{8} + \frac{1}{15} \times \frac{2}{8} = \frac{24 + 24 + 2}{120} = \frac{5}{12}.

Worth noting

13=26\frac{1}{3} = \frac{2}{6} uses bag II as it was, before the two balls were added.

Concept
A bag changed by an earlier draw
Typical time
120 s
Your time
—
The rule this testsThe theorem of total probability
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Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.

Practise this chapter →

Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.