Question

A local soccer team has 6 more games that it will play. If it wins its game this weekend, then it will play its final 5 games in the upper bracket of its league, and if it loses, then it will play its final 5 games in the lower bracket. If it plays in the upper bracket, then it will independently win each of its games in this bracket with probability 0.3, and if it plays in the lower bracket, then it will independently win each of its games with probability 0.4. If the probability that it wins its game this weekend is 0.5, what is the probability that it wins at least 3 of its final 5 games?

Answer 1

The soccer team is equally likely to be in the lower bracket and the upper bracket.

Now let the games played in upper bracket follow binomial distribution with probability of winning as 0.3

So the probability to win atleast 3 of its final games is:

P(X>=3) = 0.16308

The games played in the lower bracket also follow binomial distribution with the probability of winning 0.4.

P(X>=3) = 0.31744

Since both the venets are equally likely. The overall probability fo winning atleast 3 out of the final 5 games is:

0.5*0.16308+0.5*0.31744 = 0.24

24%