A bag has 4 red and 6 black balls. Two are drawn one after the other without replacement. What is the probability that the second ball is red?

\frac{2}{15}

Correct answerYour answer

\frac{1}{3}

Correct answerYour answer

\frac{2}{5}

Correct answerYour answer

\frac{4}{9}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

Split on the first ball: $\frac{4}{10} \times \frac{3}{9} + \frac{6}{10} \times \frac{4}{9}$ .
$\frac{12 + 24}{90} = \frac{2}{5}$ , the same as for the first ball.

Worth noting

$\frac{1}{3}$ assumes the first ball was red; it may have been black.

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

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

Explanation