You flip a coin 5 times. What is the probabilty that it will come up heads (not heads or tails, but heads specifically) AT LEAST twice? (I don't know the answer; more importantly, what is the formula?)

If I figure correctly, there are 32 possibilities (2 to the 5th power). 26 outcomes are favorable, but I'm not sure how to get the 26 without doing it manually.

You are correct. The (C)ombination formula is: nCr = n!/r!(n-r)! where n = # trials and r = # to be selected. Not sure about a shortcut but it can be solved one of two ways, how many meet your criteria or total trials minus how many do not meet your criteria. I will use option 2 since it is shorter. How many results have 0 or only one head? 5!/0!(5-0)! + 5!/1!(5-1)! = 6 that do not have at least 2 heads or 32-6 = 26 that have at least two heads. Joe.

you would use the binomial distribution function and solve like ... X = P(exatly 2) + P(exatly 3) + P(exatly 4)+P(exatly 5)

It is slightly less work to do it Joe's way: calculate Y = probability(less than 2 heads) and then X = (1 - Y) Y = P(exactly 0 heads) + P(exactly 1 heads) = (1/32) + (5/32) = (6/32) Then X = Probability(2 or more heads) = 1 - Y = (26/32) using the good old binomial probability distribution mentioned by rosy2: P(exactly 0) = (5 choose 0)*(p^0)*((1-p)^5) = 1*1*(1/2)^5 = 1/32 P(exactly 1) = (5 choose 1)*(p^1)*((1-p)^4) = 5*((1/2)^1)*(1/2^4) = 5/32

It's just binomial distribution dudes. 1 - (P(0) + P(1) ) Hmm nevermind, I didn't see it was there. you had to make it all complicated.