The probability of picking 8 left footed players at random for a soccer team of 11 where 12 percent of population is left footed

Question

I'm trying to answer this question for my son whose favourite team has 8 left footed players and he wanted to know how unusual this is. While there may be lots of ways of approaching the answer for this - including looking at numbers of left footed players in the soccer leagues compared to right footed - I thought i would use a random approach based purely on the whole population which shows 12.1% are left footed.

I think this a binomial probability question but its a long time since I did statistics and cant remember the approach and formulas.

Answer 1

Probability p=0.12. Let q=1-p=0.88.

(p+q)¹¹=p¹¹+11p¹⁰q+(11)(10)p⁹q²/2+(11)(10)(9)p⁸q³/6+...

The probability of exactly 8 out of 11 players is therefore:

(11)(10)(9)p⁸q³/6=165(0.12⁸)(0.88³)=0.000004835 (0.00048% approx).

This is 1/206832 approx.

If p=0.121, then q=0.879 and the probability is 0.000005149 (0.00051% approx). This is 1/194208 approx.

So it is highly improbable! It suggests that there is a factor other than randomness operating here. Maybe left-footers make better players!