A bag contains 5 white and 3 black balls, and 4 are successively drawn out and not replaced. What’s the chance of getting different colours alternatively?
Solution
Probability of alternate balls when first ball is black = $$\frac{5}{8}\times\ \frac{3}{7}\times\ \frac{4}{6}\times\ \frac{2}{5}=\frac{1}{14}$$
Probability of alternate balls when first ball is white = $$\frac{3}{8}\times\ \frac{5}{7}\times\ \frac{2}{6}\times\ \frac{4}{5}=\frac{1}{14}$$
Overall Probability = $$\frac{1}{14}+\ \frac{1}{14}=\frac{1}{7}$$
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