# Thread: Probability

1. ## Probability

3 dice are tossed. What is the probability that 1 was obtained on two of the dice given that the sum of the numbers on the three dice is 7?

So |S| = 6^3 = 216.

My event set will be the combinations of the numbers on the 3 dice which sum to 7. So, E = {(1,1,5), (1,2,4), (1,3,3), (2,2,3)}
My condition is F = 1 was obtained on two of the dice given they sum to 3 = {(1,1,5)}

I'm not sure if I should do P(E|F) or P(E^F) as I'm kind of lost.

Could it be p(E)*p(F) which is (4/216) * [(1/4)/216]?

2. ## Re: Probability

How many ways can you get a total of 7?
You list 4 in Set E, but
you can get two 1s and a 5 in three ways, (1, 1, 5), (1, 5, 1) and (5, 1, 1)
you can get a 1, a 2 and a 4 in six ways (1, 2, 4), (1, 4, 2), (2, 1, 4), (2, 4, 1), (4, 1, 2) and (4, 2, 1)
you can get a 1 and two 3s in three ways
you can get two 2s and a 3 in three ways

So there are 3+6+3+3 = 15 ways of getting a total of 7.

How many of these contain two 1s ? 3 ways

Can you finish it off?

3. ## Re: Probability Originally Posted by MrJank 3 dice are tossed. What is the probability that 1 was obtained on two of the dice given that the sum of the numbers on the three dice is 7?
So |S| = 6^3 = 216.
My event set will be the combinations of the numbers on the 3 dice which sum to 7. So, E = {(1,1,5), (1,2,4), (1,3,3), (2,2,3)}
My condition is F = 1 was obtained on two of the dice given they sum to 3 = {(1,1,5)}
I'm not sure if I should do P(E|F) or P(E^F) as I'm kind of lost.
Could it be p(E)*p(F) which is (4/216) * [(1/4)/216]?
Look at this expansion. In it the coefficient of each term tells tells the number of ways to get the sum equal to the exponent. The term $15x^7$ tells us that there are fifteen ways to toss three dice and get a seven.

So $\dfrac{3}{15}=\dfrac{1}{5}$ is your answer.

BTW: Using the expression $\displaystyle {\left( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^N}$ in WolframAlpha we can get the number of way to get any number when tossing $N$ dice.