MATH: AP statistics probability question on the Game of Craps?

Jack

It sounds like Binomial with n = 10 and p = 2/9

The "expected" is equal to np = (10)(2/9) = 2.2

Using the Binomial find P(X > 2.2). Since this is a "discrete" distribution, only integers are possible. So, the P(X > 2.2) = P(X ≥ 3) which is also equal to the complement = 1 - P(X ≤ 2)

1 - P(X ≤ 2) = 1 - P(0) - P(1) - P(2)

If you use the Binomial formula with n=10 and p = 2/9 ...

1 - P(0) - P(1) - P(2) = 0.3899

So, the probability that you win on the first roll more times than expected is 0.3899

Hope that helps

You are correct about the winning percentage of 2/9. So you would expect

to win (2/9) * 10 = 20/9 = 2.222... times.

So in 10 games you would win more times than expected if you win >= 3 times,

and you would lose more times than expected if you only win 0, 1 or 2 times.

So now we have a binomial distribution problem.

The probability of losing every time is (10 C 0) *(2/9)^0 * (7/9)^10.

The probability of winning exactly once is (10 C 1) *(2/9)^1 * (7/9)^9.

The probability of winning exactly twice is (10 C 2) *(2/9)^2 * (7/9)^8.

Adding these three values gives a probability of 0.61008. This is the probability

of winning fewer times than expected.

Thus the probability of winning more often than expected is 1 - 0.61008 = 0.3899,

or about 39%.

Craps Statistics

the ans u have IS the ans to ur Q