## Problem¶

4 members of Smith family and 4 of Jones family are pooled to form all possible pairs, all equally likely. Find \(N\), the number of smiths which have a smith partner.

## Solution¶

\(X= \sum_{i=1}^4 I_i\)

Where \(I_i=1\) iff \(i^{th}\) meber of smith family has some smith member as it’s partner

Number of ways \(i^{th}\) smith member can have a smith partner (Choose 1 from remaining 3 Smiths, form group of the remaining 6 ) $ = \binom{3}{1} \times \frac{6!}{(2!)^3} $

Hence, \(P(I_i=1)= \frac{\binom{3}{1}\frac{6!}{(2!)^3}}{\frac{8!}{(2!)^4}} = \frac{3}{14}\)

The trickery happens because at the next step I am going to over count, since the wording in the question is a bit ambiguous and we are concerned with the number of smiths have a partner. I should have multiplied \(P(I_i=1)\) by \(4\) and then divided by 2, but I am going to overcount with ambiguity.

\(N = 4 \times \frac{3}{14} = \frac{6}[7]\) which ofcourse is not an integer, but did we expect it to?