probability..help?

Question

Two fair dice are tossed 5 times.Find the probabilty of getting a sum of 7 on the two dice at least twice.

please show the workings...
the answer given is 0.161

Answer 1

Look at it this way... the chance of rolling a seven on two dice is 1 in 6 (show the 36 combinations and 6 are totals that add to 7).

So the chance of getting a sum of 7 on the two dice in one roll is 1/6.
The chance of getting a sum of 7 on the two dice at least twice is: 1 minus the chance of getting no sevens or exactly 1 seven.

P(no sevens) = 5/6 x 5/6 x 5/6 x 5/6 x 5/6
P(exactly one seven) = 1/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5 (ways to pick where the seven occurs).

P(no sevens) = (5^6) ^ 5 = 3125 / 7776
P(one seven) = 3125 / 7776

P(at least two sevens) = 1 - 6250 / 7776
= 1526 / 7776
= 19.6245%

Answer 2

bandf is correct, the answer is 0.1962.

furthermore ECEngr calculated the probability that the first two rolls are where the 7's occur. there are 9 other locations for the two 7's to show up (i.e. a total of 10), for a probability of 0.16075. BUT this would be the answer if the question was "...probability of EXACTLY two 7's..." you asked for AT LEAST two 7's. again bandf correctly answered your question.

Answer 3

first consider two fair dice
the no. of combination is 36
and the no. of sum which equals to 7 is 6

P(sum = 7) = 6/36 = 1/6
P(sum ≠7) = 30/36 = 5/6

next we need at least two times out of 5 tries
that sum of 7 will appear

P(two sum = 7) = (1/6)(1/6)(5/6)(5/6)(5/6)
P(two sum = 7) = 125/7776
=1.61%

Answer 4

chance of getting 7 in one throw is 1/12. in 5 throws is 5/12. getting it twice is 5/24 = 0.208 i think