Question

A bag contains $5$ white and $3$ black balls. Four balls are successive drawn out without replacement. What is the probability that they are alternately of different colours?

Answer 1

Total number of balls $=5+3=8$
Probabiliyt of getting one white ball $=\frac{5}{8}$
Number of remaining balls $=8-1=7$
out of them there are $4$ white and $3$ black balls
Probability of being black ball $=\frac{3}{7}$
Number of remaining balls $=7-1=6$ where are $4$ white and $2$ black balls
Probability of being third white ball $=\frac{4}{6}$
Number of remaining balls $=6-1=5$
where are $3$ white and $2$ black
Probability of being fourth black ball $=\frac{2}{5}$
Events are independent every time
Probability of being different colours ball
$=\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}\times 2$
$=\frac{1}{14}\times 2=\frac{1}{7}$