SOLUTION: What is the probability that 15 rolls of a fair die will show 5 threes?

Question 975208: What is the probability that 15 rolls of a fair die will show 5 threes?

Found 3 solutions by Boreal, solver91311, Edwin McCravy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
Probability of a 3 is (1/6)
This is a binomial with 15C5*(1/6)^5*(5/6)^10 = 0.0624
There are 15 rolls and 3003 different ways 5 threes can come up.
The overall probability is that time (1/6)^5 times and (5/6)^10 times.
The expected value would be 2.5, so this is reasonable.
While the numbers are too small to use the normal approximation, the variance is np(1-p)=75/36
The standard deviation would be about 1.35.
This is about 2 sd s above the mean, so again it is in the range of what one would expect.

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Exactly 5 threes:

At most 5 threes:

At least 5 threes:

John

My calculator said it, I believe it, that settles it

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!

I think what you want is a binomial probability, not what the others gave you. N = number of trials = 15 x = number of times out of those to succeed = 5 p = the probability of 1 success in 1 trial = 1/6 NCx = 15C5 = Or else on TI-83 or 84 Press 2ND, Press VARS, scroll down to highlight binompdf(, Press ENTER -------------------- If you have newer model, make it read trials: 15 p:1/6 x value:5 Paste Highlight Paste press ENTER You will see binompdf(15,1/6,5) --------------------- If you have older model, you just see binompdf( make it read binompdf(15,1/6,5) -------------------- press ENTER Read .0623715619 Edwin

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