# Probability equation

1. Jul 30, 2004

### Muck

The question: A baseball player has a 5% chance to hit a homerun each at bat. If the player is up 4 times, what is the chance he hits 2 or more homeruns. I came up with the answer, but this was a long process. I need a formula. And is there an easier way of counting the possibilities?

I did it like this:

Probability of 0 out of 4 + 1 out of 4 = 1.401875

Probability of 0 out of 4 is (0.95)(0.95)(0.95)(0.95)(1) with 1 possibility

Probability of 1 out of 4 is (0.05)(0.95)(0.95)(0.95)(4) with 4 possibilities so

The other 11 possibly scenarios are 6 to do 2 out of 4, 4 to do 3 out of 4, and 1 to do 1 out of 4, for a total of 16 (4x4) but is there an easier way than counting these?

Thank you!

EDIT: for 'player' :)

Last edited: Jul 30, 2004
2. Jul 30, 2004

### AKG

Your calculations are wrong somewhere. For one, it should be obvious that the first two cases can't have a probability greater than 1, in fact the total probabilities of all the cases should be one. The probability for the first two cases is 0.98598125. Anyways, you can see it was pretty easy to calculate the first two numbers. The probability for 2, 3, or 4 homeruns (i.e. at least 2) is 1 - 0.98598125. You could also calculate it as:

Probability of 2 out of 4 is (0.05)(0.05)(0.95)(0.95)(6) with 6 possibilities

Probability of 3 out of 4 is (0.05)(0.05)(0.05)(0.95)(4) with 4 possibilities

Probability of 4 out of 4 is (0.05)(0.05)(0.05)(0.05)(1) with 1 possibility

Sum those together you'll get the same number as 1 - 0.98598125.

3. Jul 30, 2004

### HallsofIvy

I wasn't aware that it was the baseball that hit the homerun!

Assuming that any batter has probabilty 0.05 of hitting a homerun at any "at bat", then the probability of a batter hitting k homeruns in n "at bat"s is nCk(0.05)k[/sub](0.95)n-k where nCk is the binomial coefficient: n!/(k!(n-k)!).

4. Jul 30, 2004

### Muck

Thank you.

5. Aug 15, 2004

### robert Ihnot

This is actually a Bernoulli trial problem that can be seen as tossing a coin, where the chance for heads is .05. Since the total probability is 1 = (H+T)^4. We can just look at 1 minus the tail of the series:

1-4HxT^3-T^4.