Question

Two balls are drawn without replacement from an urn containing 10 black balls and 5 white balls. Given that the first ball is black, what is \[P(F\mid E)\] when \[F\] is “second ball is black”?

Options

• \[\frac{10}{14}\]

• \[\frac{10}{15}\]

• \[\frac{9}{15}\]

• \[\frac{9}{14}\]

Answer 1

After one black ball is removed, 9 black balls remain. The urn then contains 14 balls, so \[P(F\mid E)=\frac{9}{14}\].