binomial probabilities

Question

A high school baseball player has a 0.163 batting average. In one game, he gets 9 at bat. What is the probability he will get at least 6 hits in the game? Round your answer to four places.

Answer 1

Hi Alma,

The formula you need to solve this problem is:

P(X=a)=C(n,x)p^xq^n-x

Let’s break this down:

P=Probability, essentially what we are looking for

X=a=Number of successes in binomial outcome

n=number of trials

p=probability of success

q=probability of failure

That’s the general idea. Now, breaking it down specifically for this problem:

X=a=6 This is the number of hits we are looking for.

n=9, the number of at-bats

p=0.163, the player’s batting average, the probability that he gets a hit on any given at-bat

q=0.837, computed as 1-p, the probability that he does not get a hit

Now, we can substitute these values into the initial formula:

P(X=6)=C(9,6)*0.163⁶*0.837^9-6

C(9,6)=9!/(6!3!)

If you haven’t learned this yet, !=factorial, which essentially means take 9*8*7*6..all the way down to 1. You can plug that into a TI-83 Plus or greater calculator, but I can show you how to compute by hand if necessary.

Anyway,

C(9,6)=84

Substituting and simplifying:

P(X=6)=84*0.163⁶*0.837³

P(X=6)=0.0009

But we’re not done yet. Recall that the question asked you for AT LEAST 6 hits. That means we have to compute probabilities for 7, 8, and 9 as well. Ultimately, we add them all together: I will leave the probability calculations for P(X=7,8,9)to you. Use the same formula we used for 6 and change the appropriate values, Ultimately, we must add:

P(X=6) + P(X=7) + P(X=8) + P(X=9)=

0.00098

This is extremely low, which makes sense because that batting average is rather low. It is unlikely that a player with a 0.163 batting average would get 6 hits in 9 at-bats. I hope this helps.