Question

Two balls are drawn successively without replacement from a box which contains 4 white balls and 3 red balls. Find the probability that

(a) the first ball drawn is white and the second is red;
(b) both balls are red.

Answer 1

(a) The second event is dependent on the first.
\[
P\left(E_{1}\right)=P(\text { white })=\frac{4}{7}
\]
There are 6 balls left and out of those 6 , three of them are red. So the probability that the second one is red is given by:
\[
P\left(E_{2} \mid E_{1}\right)=P(\text { red })=\frac{3}{6}=\frac{1}{2}
\]
Dependent events, so
\[
P\left(E_{1} \text { and } E_{2}\right)=P\left(E_{1}\right) \times P\left(E_{2} \mid E_{1}\right)=\frac{4}{7} \times \frac{1}{2}=\frac{2}{7}
\]
(b) Also dependent events. Using similar reasoning, but realising there will be 2 red balls on the second draw, we have:
\[
P(R R)=\frac{3}{7} \times \frac{2}{6}=\frac{1}{7}
\]