Binomial Maths Problem...

jarvisob

Hi all just looking for some help with this probability problem....

Two dice are thrown together. Determine the probability that the sum of the two numbers is 8. The dice are thrown 12 times. Determine the prob that a sum of 8 occurs (1) three times (2) never (3) at least 4 times.

So the binomial formula is



So for first part i know n=12 and k=3. The prob of 2 dice equaling 8 is 5/36.
Putting these values into equation I worked the final answer out to be 15.31%.


For second part i worked it out to be 16.62%


I'm stuck with the final part. What value to i put in for k and what will be my binomial coefficient? Any help would be greatly appreciated. Test tomorrow! Cheers

RoundTower

The best way to get an exact answer is to find the probability an 8 occurs 0, 1, 2 or 3 times and take that away from 1. You've already done it for 0 and 3, so now do it for 1 and 2.

RoundTower

if you have a fancy tables book or a fancy calculator you might be able to get a result more straightforwardly, I can't see how to do it on Wolfram Alpha

ArmCandyBaby

Yeah for complicated questions you'd use tables or a normal approximation. I think this is the Wolfram anwser:

http://www.wolframalpha.com/input/?i=CDF[BinomialDistribution[12%2C+0.138888889]%2C+4]

jarvisob

Thanks very much people

dafunk

I would approach it as round tower suggested:

P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12) = 1

P(At least 4) = P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)+P(11)+P(12)

= 1 - [P(0)+P(1)+P(2)+P(3)]