Assertion (A): Bag I has 1 red and 4 black balls, and bag II has 3 red and 2 black balls. A bag is picked at random and a red ball is drawn from it. Then bag II is three times as likely as bag I to have been picked.
Reason (R): When $P(\text{I}) = P(\text{II})$ , $\frac{P(\text{II} \mid R)}{P(\text{I} \mid R)} = \frac{P(R \mid \text{II})}{P(R \mid \text{I})}$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

R: by Bayes' theorem both posterior chances have the same denominator $P(R)$ , and the equal $P(\text{I})$ and $P(\text{II})$ cancel, leaving the ratio of $P(R \mid \text{II})$ to $P(R \mid \text{I})$ . True.
A: that ratio is $\frac{3/5}{1/5} = 3$ . True, and R is the reason.

Worth noting

When the bags are equally likely to be picked, each bag's chance after a red ball is in proportion to its chance of giving red.

Concept: Bayes' theorem
Typical timeTypical: 90 s
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The rule this testsBayes' theorem

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

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