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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.
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(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:
$P(R R)=\frac{3}{7} \times \frac{2}{6}=\frac{1}{7}$
by Diamond (44,867 points)

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