Players A and B play a basketball game in which they take turns shooting the ball, and the first player to make their shot wins. Player A has probability 2/3 of making each of her shots. Player B has probability 1/2 of making each of his shots. If Player A shoots first, what is the probability that she will win

Player A has a probability 2/3 of making each of her shots, then she has a probability 1/3 of missing each shot.

Player B has a probability 1/2 of making each of his shots, then he also has a probability 1/2 of missing each shot.

Let's separate each case.

Let's define:

P(x) = probability of winning at the "x" shot.

Player A wins on the first shot.

Because she has a probability 2/3 of making each of her shots, the probability of winning at the first shot is

P(1) = 2/3

Now let's see the next case, player A wins at her second shot.

This means that first, she misses her first shot, with a probability of:

p₁ = 1/3

Player B must miss his shot, the probability is:

p₂ = 1/2

Now player A must make her shot, so the probability is:

p₃ = 2/3

The joint probability is the product of the individual probabilities, so we have:

P(2) = (1/3)*(1/2)*(2/3) = 1/9

Now we can see the pattern, for P(3) we have

A misses: p₁ = 1/3    (first shot of A)

B misses: p₂ = 1/2

A misses: p₃ = 1/3   (Second shot of A)

B misses: p₄ = 1/2

A makes the shot: p₅ = 2/3

P(3) =  (1/3)*(1/2)*(1/3)*(1/2)*(2/3) = 1/54

We already can see the pattern.

P(n) = (1/3)^(n - 1)*(1/2)*(n - 1)*(2/3)

Player A has a probability P of winning, and we can write P as:

P = P(1) + P(2) + P(3) + ...

Then we will have:

P = 2/3 + 1/9 + 1/54 + 1/324 + ... ≈ 0.8

17.08

correct me if i am wrong.

yes

step-by-step explanation:

they have to be opposite and reciprocal of eachother.