A bag contains 6 blue and 6 green balls.

Question

A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is :

Accepted Answer

We want every pair drawn to have one blue and one green. Since there are 6 blue and 6 green balls, we will draw a total of 6 pairs.

For each pair to be mixed (one of each colour), we must match the 6 blue balls with the 6 green balls in pairs. The number of such ``good'' pairings is counted by choosing and matching one ball of one colour with one ball of the other colour.

The total number of ways to split 12 balls into 6 pairs (without caring about the order of drawing inside a pair) is counted separately. The required probability is:

Probability $=\frac{\lfloor 6	imes \lfloor 6}{(\frac{12}{(\lfloor 2{)}^{6}\lfloor 6})	imes \lfloor 6}=\frac{16}{231}$

So, the probability that each drawn pair consists of one blue and one green ball is $\frac{16}{231}$.

Solutions · 1