# Rolling 2 fair, 6 sided dice find P(sum=12, 2times in 36 rolls)

I have a dice problem. If we roll 2 fair dice, and the sum is 12 then our test is a pass, otherwise its a fail. What is the probability that the number of passes in 36 tests is greater then 1.

S={1,2,3,4,5,6}

(S,S) = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

event

Let e = sum of both dice is 12 e in s = {(6,6)} P(e)=1/36

Then, i guess i would need to take the inverse of the failures

1 - (35/36)$*$(35/36)$*$(35/36)$*$ ... $*$(35/36)

1 - .3267 = .6372

If im looking at this right, then i have just calculated the probability that my test passes at least once. How do i modify this to calculate at least 2 times?

EDIT:

I think i should look at it like this: P(eefff...f) = 1/36^2 * 35/36^34

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The complement of "greater than one" is "none or exactly one". You've found the probability of none. Now find the probability that exactly one is a pass. Then add the two together and subtract from $1$. – David Mitra Feb 25 '13 at 20:34

Note that there is only way way to achieve a $12$ when rolling two dice (whose outcomes we will denote by the random variables $X_{1}$ and $X_{2}$) and taking their sum, which is when both dice show $6$, which are both independent events with probability of $\frac{1}{6}$, therefore:

$$P(X_{1}+X_{2}=12)=P(X_{1}=6 \cap X_{2}=6)=P(X_{1}=6)P(X_{2}=6)=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$

We now know the outcome of each Bernoulli trial is a success with probability $\frac{1}{36}$, therefore, we can model the probability of the number of successes in $36$ trials using a Binomial distributions $Y\sim B(36,\frac{1}{36})$. Therefore, the probability of achieving exactly $2$ successes is:

$$P(Y=2)={36\choose 2}\left(\frac{1}{36}\right)^{2}\left(\frac{35}{36}\right)^{34}\approx0.1865$$

EDIT: Your question asks for the probability that there are at least $2$ successes (i.e. $P(Y\geq2)$), which is:

$$P(Y\geq2)=1-P(Y\leq1)=1-P(Y=0\cup Y=1)=1-\left(\left(\frac{35}{36}\right)^{36}+{36 \choose 1}\left(\frac{1}{35}\right)\left(\frac{35}{36}\right)^{35}\right)\approx1-0.2642\approx0.7358$$

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Thanks!! i picked up on that almost as soon as i clicked the post button. can you just tell me what you do the 36 pick 2? I guess its because it doesnt matter exactly where the 2 passes are. – KevinCameron1337 Feb 25 '13 at 20:32
@Special--k Exactly, it is because we want $2$ successes (hence $\left(\frac{1}{36}\right)^{2}$) and $34$ failures (hence $\left(\frac{35}{36}\right)^{34}$) and they can be anywhere in the chain of trials (so there are ${36 \choose 2}$ ways of achieving this). – Shaktal Feb 25 '13 at 20:34
Note the problem is to find the probability that there are at least two "passes", not exactly two. – David Mitra Feb 25 '13 at 20:36
so the 0.18 is the probability of exactly 2. or Greater then 1? I dont think this is going to work, because i need to know greater then 1. Where i think this formula provides exactly 2. – KevinCameron1337 Feb 25 '13 at 20:37
@DavidMitra Ahh damn, I didn't read the question fully! Thank you for pointing that out! – Shaktal Feb 25 '13 at 20:37