Question

The probability that a professional baseball player will get a hit is 1/3. Calculate the exact probability that he will get at least 3 hits in 5 attempts.

Answer 1

you know about binomial probability??

Answer 2

no?

Answer 3

oh!!

Answer 4

yea

Answer 5

At least 3 means 3 or more, so 3, 4, or 5. We can calculate the probablity that he gets 3 hits, 4hits, and 5 hits seperately and then add 'em all up.
3 hits:
(1/3)(1/3)(1/3)(2/3)(2/3)
4 hits:
(1/3)(1/3)(1/3)(1/3)(2/3)
5 hits:
(1/3)(1/3)(1/3)(1/3)(1/3)

Answer 6

Oo .. @SmoothMath i think quite not.
wolframalpah is giving me funny results, you do know how to type C
http://www.wolframalpha.com/input/?i=%285C3%29+*+%281%2F3%29%5E3+*+%281%2F2%29%5E2+%2B+%285C4%29+*+%281%2F3%29%5E4+*+%282%2F3%29+%2B+%285C5%29+*+%281%2F3%29%5E5

Answer 7

Ohright... doesn't matter what order.

Answer 8

Sooo:
3 hits:
(1/3)^3(2/3)^2 * nCr(5,2)
4 hits:
(1/3)^4(2/3) * nCr(5,1)
5 hits:
(1/3)^5

Answer 9

http://www.wolframalpha.com/input/?i=%285+choose+3%29+*+%281%2F3%29%5E3+*+%281%2F2%29%5E2+%2B+%285+choose+4%29+*+%281%2F3%29%5E4+*+%282%2F3%29+%2B+%285+choose+5%29+*+%281%2F3%29%5E5

Answer 10

so is the answer 51/243?

Answer 11

Experiment, Alpha accepts "(5 choose 2)"

Answer 12

http://www.wolframalpha.com/input/?i=%285Combination3%29+*+%281%2F3%29%5E3+*+%281%2F2%29%5E2+%2B+%285Combination4%29+*+%281%2F3%29%5E4+*+%282%2F3%29+%2B+%285Combination5%29+*+%281%2F3%29%5E5

Haha .. found other way. I think this type of problem is handled with binomial probability. Though I can never be sure.
http://en.wikipedia.org/wiki/Binomial_probability

Answer 13

Okay, Jerica, let's look at the probability of a SPECIFIC batting event to happen, like:
hit, hit, hit, miss, hit
The probability of that is:
(1/3)(1/3)(1/3)(2/3)(1/3)
right? Or I can rewrite that as:
(1/3)^4(2/3)

Answer 14

yea

Answer 15

Okay, cool. So for a batting event with 4 hits involved, I get (1/3)^4(2/3)

Answer 16

but how many possible such events are there? Well, I have:
hhhhm
hhhmh
hhmhh
hmhhh
mhhhh
so, five.
Therefore 5*(1/3)^4(2/3) tells me the probability of getting 4 hits.

Answer 17

@SmoothMath continue ...

Answer 18

To get 5 hits, I get (1/3)^5, and there's only one possible way to get 5 hits:
hhhhh,
so I don't need to multiply by any new number.

Answer 19

To get 3 hits is a little tricky, and it helps to understand a cool tool called combinations. Do you know what I mean if I say "5 choose 2" or nCr(5,2) or \[\left(\begin{matrix}5 \\ 2\end{matrix}\right)\]

(These are all notations for the same thing)

Answer 20

ok I get where your going

Answer 21

Yeah? To get 3 hits, I get (1/3)^3*(2/3)^2
That's the probability of an event like:
hhhmm
But then I need to count the number of events LIKE that, that is, with two hits. The number of events like that is 5 choose 2 because I have 5 at bats and I am choosing two of them to miss on.

Answer 22

(Btw, if you prefer to think of it as 5 choose 3, that is fine also. Choosing 3 to hit on is the same as choosing 2 to miss on, and they give me the same answer, which is 10.

Answer 23

but my teacher got the answer 51/243
I just don't know how?

Answer 24

calculating mistake
http://www.wolframalpha.com/input/?i=%285Combination3%29+*+%281%2F3%29%5E3+*+%282%2F3%29%5E2+%2B+%285Combination4%29+*+%281%2F3%29%5E4+*+%282%2F3%29+%2B+%285Combination5%29+*+%281%2F3%29%5E5

Answer 25

It may prove to be pure hubris, but I stubbornly claim that your teacher is incorrect and I am correct. =D

Answer 26

I think you should add combination to getting exactly 3 hits. I'm sure you will get the exact result.

Answer 27

ok thanks @SmoothMath & @experimentX

Answer 28

Haha okay so it did prove to be pure hubris.

Answer 29

Method was correct. Made a mistake inputting into alpha. Had a (1/2) somewhere where I should have had a (2/3).
The answer comes out to be, like experiment's previous link shows, 17/81, which is just a reduction of 51/243, the answer your teacher gave.

Answer 30

Oo .. I take my word back.