Probability/Die question.

jmsal · #1 $Old$ May 28th, 2009, 04:23 PM

In rolling two fair dice simultaneously, the probability of rolling a 7 or 11 is 2/9. What is the approximate probability of rolling a 7 or 11 two or more times in ten rolls of two dice?

Need some direction I really don't know where to start with this one.

artvandalay11 · #2 $Old$ May 28th, 2009, 04:42 PM

The probability of rolling 7 or 11 2 or more times in 10 rolls P $(7\vee 11\geq 2)$ is the same as 1-probability of rolling 7 or 11 less than 2 times in 10 rolls.

This is equal to 1-(probability of rolling 7 or 11 zero times +probability of rolling 7 or 11 one time)

The binomial distribution is the best tool to compute the probability of rolling 7 or 11 x times

Probabilitiy of rolling a 7 or 11 one time out of 10 rolls= $\binom{10}{1}\left(\frac{2}{9}\right)^1\left(1-\frac{2}{9}\right)^9=\binom{10}{1}\frac{2}{9}\left(\frac{7}{9}\right)^9$

and the probability of rolling a 7 or 11 zero times out of 10 rolls= $\binom{10}{0}\left(\frac{2}{9}\right)^0\left(1-\frac{2}{9}\right)^{10}=\binom{10}{0}\left(\frac{7}{9}\right)^{10}$

Search google for the binomial distribution if you're still confused

jmsal · #3 $Old$ May 28th, 2009, 04:48 PM

Thank you so much. I forgot to mention that this is a multiple choice:
A:0.08 B:0.23 C:.30 D:0.69 E:.77.

Going from youre calculations the best answer is A:0.08

artvandalay11 · #4 $Old$ May 28th, 2009, 04:51 PM

no the best answer is D

add the two calculations, subtract that from one, and you get .687