Die throwing probability question

[elmarsur]

Homework Statement

Problem:
Suppose in a certain game a player receives $1 for each "five"
and each "six" thrown on a single die.
a. If the die is thrown six times, find the mean and variance of
the amount the player receives.
b. If the player throws repeatedly until he gets $10, find the
mean of the number of throws necessary

Homework Equations

The Attempt at a Solution

Attempt:
I think this is binomial. Then:
1a)
n=6
p=2/6 = 1/3
mean = n*p = 2
variance= n*p*(1-p) = 6*1/3*2/3 = 1/3
Is this correct?

1b)
If $10 is the amount gained,it means:
x=10 (successes)
n=trials
p=1/3
1-p=2/3

=> {n!/[x!(n-x)!]}*(p^10)*(1-p)^(n-10) ?
Is this correct, although tedious?

Thank you very much for any help.

 

[mighty2000]

Hello, thank you for your post. Your attempt at solving the problem is correct. This is indeed a binomial distribution, where each throw is a success if a "five" or "six" is rolled.

For part a), the mean and variance are calculated correctly using the formula for a binomial distribution.

For part b), your approach is correct but may be a bit tedious. Another approach to solving this problem would be to use the geometric distribution, since we are looking for the number of trials until a certain amount is reached. The mean of a geometric distribution is equal to 1/p, where p is the probability of success. So in this case, the mean would be 3 throws until the player reaches $10. However, your approach using the binomial distribution is also valid.

I hope this helps! Let me know if you have any further questions.