Question

Suppose the probability that a softball player gets a hit in any single at-bat is 0.300. Assuming that her chance of getting a hit on a particular time at bat is independent of her other times at bat, what is the probability that she will not get a hit until her fourth time at bat in a game

Answer 1

Answer:

$P(X=4)=(1-0.3)^{4-1} 0.3 = 0.103$

The probability that she will not get a hit until her fourth time at bat in a game is 0.103

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Solution to the problem

For this case we want this probability

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(1-0.3)^{4-1} 0.3 = 0.103$

The probability that she will not get a hit until her fourth time at bat in a game is 0.103