Question

A bag contains 4 white and 2 black balls 4 balls are successively drawn out without replacement.what is the probability that they are alternatively of different colour?

Answer 1

Given :

No of white balls in the bag= 4

No of black balls in the bag = 2

To Find :

If 4 balls are successively drawn out without replacement , then the probability that they are alternatively of different colors = ?

Solution :

Here the two favourable  conditions possible are :

$E_1:WBWB$  or   $E_2:BWBW$

Here B denotes black ball and W denotes white ball .

So,

$P(E_1) = P(W) \times P(\frac{B}{W}) \times P(\frac{W}{W\cap B}) \times P(\frac{B}{W\cap B \cap W})$

           = $\frac{4}{6} \times \frac{2}{5} \times \frac{3}{4} \times \frac{1}{3}$

            $= \frac{1}{15}$

Similarly ,

$P(E_2) = \frac{2}{6} \times \frac{4}{5} \times \frac{1}{4} \times \frac{3}{3}$

          = $\frac{1}{15}$

So, total probability  for the event is :

= $P(E_1) + P(E_2)$

=$\frac{1}{15} + \frac{1}{15}$

=$\frac{2}{15}$