A pair of fair dice is tossed 3 times.
Find the probability a seven appears 3 times.

Question

anonymous · Answer

i'm gonna go with 16.67%^3=0.00463241...

anonymous · Answer

probability that seven appears on one throw = 6/36 = 1/6

probability that it appears 3 times = 1/6 *1/6*1/6 = 1/216

anonymous · Answer

So, as a percentage, 0.463%

anonymous · Answer

thats correct but what happens when if they ask you to  Find the probability a 7 or 11 appears at least twice.

anonymous · Answer

it is a binomial probability...

probability that 7 or 11 appear on one throw = 8/36 = 2/9

probability that 7or 11 appear at least 2 times = 3C2 * (2/9)^2 * (7/9)

anonymous · Answer

P(7) =0 unless you're looking at the probability that  the sum is 7...but a fair die has numbers 1 to 6 only...no 7...thus P(7) = 0 for each toss...and thus 0 overall...

anonymous · Answer

GODinme it is the sum of 7 or 11

anonymous · Answer

sorry, it is probability that 7 or 11 appear exactly 2 times.

for at least 2 times please add this  also...
 3C3 *(2/9)^3

anonymous · Answer

ok...that wasn't in the question...who's asking the question?

anonymous · Answer

7 or 11? Thats a Craps problem...Google will help...

anonymous · Answer

heartslayer20   asked it..........

anonymous · Answer

so how do you know what it's sposed to be?

anonymous · Answer

it was a two part question you know it is a sum because of course there isnt a 7 or a 11.  The question was a two parts.  I got the first answer from the first two replies.  the second part is "A pair of fair dice is tossed 3 times.Find the probability a 7 or 11 appears at least twice.

anonymous · Answer

probability that 7or 11 appear at at least 2 times 
= 3C2 * (2/9)^2 * (7/9)    +    3C3 *(2/9)^3

anonymous · Answer

probability of a 7, 1/6th probability of an 11, 1/18th. Multiply ad nauseum...

anonymous · Answer

ok...then it can be represented as a binomial distribution:
n=3, P(7) = 1/6 for any throw...and each throw is independent...
thus: 3C3*(1/6)^3*(5/6)^0 + 3C2*(1/6)^2*(5/6)^1
P(11) = 1/18
so similarly: 3C3*(1/12)^3 + 3C2*(1/12)^2*(11/12)^1
the sum ofr P(7) + P(11) is what is wanted...

anonymous · Answer

sorry, your solution is wrong...
probability that 7 or 11 appear on one throw of a pair of dice =  1/6 + 2/36 = 8/36 = 2/9
P(AUB) = P(A) + P(B) 
here P(A n B) = 0

anonymous · Answer

i think you're right...i approached it from a different angle but i think i messed it up that way...thanks...

anonymous · Answer

ok....