Probability question

frozenclaw

Canada410 Posts

May 30 2008 14:52 GMT

Hi guys, I have a math question that's been bugging me. If a baseball player has a 1/4 chance of making a hit on any trial, what is the theoretical probability that he will make 3 or more hits in 5 trials? All I managed to determine was that there are 1024 possibilities since 1/4^5=1/1024. Alright thanks a lot guys.

goldrush

Canada709 Posts

May 30 2008 15:11 GMT

Binomial series:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

It's been a while since I did probability, but it should be:

10 * (0.25)^3(0.75)^2 [three hits] + 5 * (0.25)^4(0.75) [four hits] + 1 * 0.25^5 [five hits]
= 0.1035 or approximately 10.35%

Something like that. =P

There are only 32 different cases. Imagine that 1 represent a hit and 0 represents a miss:

11111

In reality, each number has only two cases, either a 1 or a 0. Therefore, the number of different cases is 2^5 or 32.

Myrmidon

United States9452 Posts

May 30 2008 15:18 GMT

(5 choose 3) ways to get 3 hits exactly = 10
(5 choose 4) ways to get 4 hits exactly = 5
(5 choose 5) ways to get 5 hits exactly = 1

Probability of getting a specific combination of 3 hits = (1/4)^3 * (3/4)^2 = 9/1024
Probability of getting a specific combination of 4 hits = (1/4)^4 * (3/4)^1 = 3/1024
Probability of getting a specific combination of 3 hits = (1/4)^5 * (3/4)^0 = 1/1024

By specific combination of 3 hits I mean:
hit hit hit miss miss
or
hit miss miss hit hit
etc.
(there were 10 of those as calculated at top)

Total probability:
ways to get 3 hits * prob of specific 3 hits + ways to get 4 hits * prob of 4 hits + ways to get 5 hits * prob of 5 hits

= (10)(9/1024) + (5)(3/1024) + (1)(1/1024) = 53/512 ~= 10.35%

edit: dammit my connection went down before I could post >_<;;