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

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6