We roll a fair 6-sided die 5 times. What is the probability that we get a 6 in at most 2 of the rolls

Can anyone solve this problem? It's just a basic probability problem. Any help would be appreciated! Thank you! :D

you have a 1/6 probability of rolling a 6
you have a 5/6 probability of rolling anything else

at most 2 sixes in 5 rolls means:
0 or 1 or 2 sixes

P(0 sixes) = (5/6)^5
P(1 six) = 5C1 * (1/6) * (5/6)^4
P(2 sixes) = 5C2 * (1/6)^2 * (5/6)^3

P(x <= 2) = P(0) + P(1) + P(2)

P(x <= 2) = 0.96450617283950617283950617283951
96.45% probability of at most 2 sixes in 5 rolls

Whoohoo2 shouldn't be answering probability questions here. Intuitively, his/her answer can't be right. When rolling a die five times, you are EASILY going to roll a six zero, once, or twice MANY more times than just 2.77% (1/36).

I agree with David. At most, at most 2 sixes in five rolls means: 0 or 1 or 2 sixes.

I wrote a computer program to simulate this scenario, and ran it 100 million times. The percentage I came up with is 96.44%. When rolling a die five times, 96.44% of the time a 6 was rolled exactly once or twice, or not at all.

total number of cases = 6^5=7776
required probability = 1/7776 +2/7776 =3/7776 =1/2592

To get a 6 no more than twice, you'd have to roll other than a 6 three times. The probability of other than a 6 is 5/6. other than an 6 three times is (5/6)^3 ≈ .5787

The probability of a 6 no more than twice in 5 rolls = 0.5787 = 57.87%

You're going to get a lot of answers for this, but this is the right answer: 1/36. This is backed up by the probability formula.