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xotica · Dec 14, 2008, 07:56 PM

Assume that the Navy football team has a 60% chance of beating the Army football team every time they play, and that the result of each game is independent of the results of all the other games (and assume there can be no tie games).

If the Army and Navy play 10 games, what is the probability that the Navy wins exactly 8 games?

If the Army and Navy play 100 games, what is the approximate probability that the Army win 50 or more games?

v1033 · Dec 15, 2008, 12:14 PM

What level are you working at?

Do you know how to use the binomial distribution?

ebaines · Dec 15, 2008, 12:20 PM

These are binomial distribution problems. If you are getting stuck, please tell us how you are attacking these problems, and where you are getting stuck, and then I'm sure someone will help you out.

xotica · Dec 15, 2008, 04:47 PM

I believe we did a little with distribution but not sure what the formula is to solve this problem that is where I am stuck!

galactus · Dec 15, 2008, 06:04 PM

Originally Posted by xotica

Assume that the Navy football team has a 60% chance of beating the Army football team every time they play, and that the result of each game is independent of the results of all the other games (and assume there can be no tie games).

If the Army and Navy play 10 games, what is the probability that the Navy wins exactly 8 games?

$C(10,8)(\frac{3}{5})^{8}(\frac{2}{5})^{2}$

If the Army and Navy play 100 games, what is the approximate probability that the Army win 50 or more games?

$\sum_{k=50}^{100}C(100,k)(\frac{3}{5})^{k}(\frac{2 }{5})^{100-k}$

Since this one is large, we can use the short cut method, which is close.

They allude to this method in the problem statement.

First find the mean, ${\mu}=np=100(\frac{3}{5})=60$

Next, find the standard deviation, $\sqrt{npq}=100(\frac{3}{5})(\frac{2}{5})=2\sqrt{6}$

Now, use the normal distribution using the continuity correction. Which means we subtract .5 from 50. $\frac{x-{\mu}}{\sigma}=\frac{49.5-60}{2\sqrt{6}}=-2.14$

Look this z score up in the table and subtract that value from 1. It will be very close to the value gotten from the summation method above.

In these days of tech, it is no big deal to do the first one.

xotica · Dec 15, 2008, 10:53 PM

I had solved it using the summation but instead got 0.0271... is that wrong?

ebaines · Dec 16, 2008, 06:27 AM

Originally Posted by xotica

I had solved it using the summation but instead got 0.0271...is that wrong?

0.0271 is correct.