# Binomial distribution help

1. Aug 10, 2007

### chemzz

1. The problem statement, all variables and given/known data

Gerry Infeild has batting average of 0.326 what is the probability that he will have two hits in his next 5 times at bat?Give your answer correct to 3 decimal places.

3. The attempt at a solution

It is a binomial distribution with n trails and x sucesses (hence, n-x failures) with each sucess having a probability p and each failure having a probability q.

so, p(x) = (nCx)(p^x)(q^n-x) where q=1-P

Therefore 5 trials and 2 hits hence 3 failures.

P(x) = (5C2)(2/5)^2(3/5)^2

= 30 x 4/25 x 27/125
= 3240/3125
= 1.036

is that right? Then what about his batting average?? Where do we use that?

2. Aug 10, 2007

### malawi_glenn

Batting average is how many hit per try; 0.326, means that the probabillity that he hits is 0.326

When you get a value of the probability that is larger than 1, there should be bells ringing that you have done something wrong.

3. Aug 10, 2007

### chemzz

so then what should i do? how am i supposed to do this?

4. Aug 11, 2007

### malawi_glenn

What do you think?
The probability that he hits is 0.326
What is then the probability that he misses?

What are the p, and q in the biomial-formula?

We must consider this case: Assume the probability that a certain event A came out in a single trial is p. If n independent trials are made and x is the number of times A occur. Then x is binomial distributed. And your formula seems to be the right one, but you do seem to know what to do with the 0.326

I said that this is the probability that he hits, have you tried that yet??

Last edited: Aug 11, 2007
5. Aug 11, 2007

### olgranpappy

Counting and probability can be confusing. I always like to start off by warming up a little. For example, you could ask yourself a simpler question like, what is the probability that he gets 5 hit in 5 at bats... (0.326)^5, right? since he has to hit the first time AND the second time AND the third time AND the fourth time AND the fifth time.

Or, similarly, what is the probability that he misses all 5 times... (0.326)^5 again.

It also helps to make a chart--like with 5 columns and a bunch of row--to write down all the possibilities... It might look like:

HHHHH
HHHHM
HHHMH
HHMHH
...
...
MMMMM

Of course, there are usually too many possibilities to write them all down quickly. But that is where the counting tricks you learned come in... I'll get back to that in a second. But first, a question to lead you to your answer:

What is the probability that he gets a hit the first two at bats and a miss the last three? I.e., the probability of HHMMM?

6. Aug 11, 2007

### chemzz

i did that and i get this:

=(5C2)(0.326)^2(0.674)^3

= 10 x 0.106276 x 0.454276

= 0.48

is that right!

7. Aug 11, 2007

### malawi_glenn

no, you have done one error:

(5C2)(0.326)^2(0.674)^3 = 10*0.106276*0.3062 = 0.3254
{you only took (0.674)^2}

Ergo: it is 32.5% probablity that he gets two hits in five tries.

Do you understand now WHY we got this result? And why your first attempt was wrong?

8. Aug 11, 2007

### chemzz

yup i understand it now...thnx for the correction:)

9. Aug 11, 2007

### malawi_glenn

Great! Now practice more on other problems and Good luck!

10. Aug 11, 2007

### dimensionless

The above gave the total number of possible sequences as

C(5,2) = 10

Why isn't it

C(5,2) = 5*4 = 20

?

11. Aug 11, 2007

### chemzz

coz its 5!/3!2! which iz 5 x 4 x 3 x 2 x 1/ (3 x 2 x 1)(2 x 1)

= 5 x 4 /2 x 1
= 20/2
= 10

12. Aug 11, 2007

### malawi_glenn

A quicker way is to rewrite the nominator a bit, useful when we have like C(14,8) and so on.

$$\dfrac{5!}{2!3!} = \dfrac{3!*4*5}{2!3!} = \dfrac{4*5}{2!} =10$$

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