Question

The coach of a basketball team has a bag with 20 basketballs in it, and he says there is a 75% chance that the team's lucky ball is in the bag, and a 25% chance that he accidentally left it at home. He passes the basketballs out to the team's players, including John. All the players check their basketballs but don't find the lucky ball except for John. What is the probability that John has the lucky ball?

Answer 1

Explanation:

Let L denote the event that the lucky ball is in the bag, and J denote the event that John has the lucky ball.

We want to find the probability of J given that all the other players do not have the lucky ball, which we can calculate using Bayes' theorem:

P(J | not(L except J)) = P(not(L except J) | J) * P(J) / P(not(L except J))

We know that P(J) = P(L) = 0.75, and P(not(L except J)) = P(not(L) and not(J)) + P(J), since there are only two possibilities: either John has the lucky ball, or he does not and nobody else does.

So we have:

P(not(L except J)) = (0.25 * 19/20) + 0.75*(1/20) = 0.0275

Now we need to find P(not(L except J) | J), the probability that all the other players do not have the lucky ball given that John does. Since John has the lucky ball, there are only 19 basketballs left to distribute among the other players, none of which is the lucky ball. The probability of this happening is:

P(not(L except J) | J) = (19/19) * (18/19) * ... * (1/2) * (1/1) = 1/19!

Putting it all together, we get:

P(J | not(L except J)) = (1/19!) * 0.75 / 0.0275 ≈ 0.016

Therefore, the probability that John has the lucky ball given that nobody else does is approximately 0.016, or about 1.6%.