Bag I has 3 red and 4 black balls; bag II has 5 red and 4 black. A bag is picked at random and a ball drawn from it. What is the probability that the ball is red?

\frac{8}{63}

Correct answerYour answer

\frac{3}{7}

Correct answerYour answer

\frac{31}{63}

Correct answerYour answer

\frac{1}{2}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$P(R) = P(\text{I})P(R \mid \text{I}) + P(\text{II})P(R \mid \text{II}) = \frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{5}{9}$ .
$\frac{3}{14} + \frac{5}{18} = \frac{27 + 35}{126} = \frac{31}{63}$ .

Worth noting

$\frac{1}{2}$ , 8 red balls of 16, pools the balls of both bags, but the bags hold different numbers of balls and are equally likely to be picked.

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

Explanation