Question

Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a week, what's the probability that he gets greater than 9 hits?

Answer 1

`P(X>9) = 0.3593`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=25, p=0.3089)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

For this case we want this probability:

`P(X >9)`

And we can use the complement rule like this:

`P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)]`And we can find the individual probabilities like this:

`P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^(25-0) =0.0000974`

`P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^(25-1) =0.0011`

`P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^(25-2)=0.00584`

`P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^(25-3)= 0.02`

`P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^(25-4)=0.049`

`P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^(25-5)=0.092`

`P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^(25-6) = 0.138`

`P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^(25-7)=0.167`

`P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^(25-8)=0.168`

And in order to do the operations we can use the following excel code:

"=1-BINOM.DIST(8,25,0.3089,TRUE)"

And we got:

`P(X>9) = 0.3593`