Thread: Combinatorics/Probability task

call the prizes x an y and the family members a, b, c. the total number of ways to match family members with prizes is ax, bx, cx ay, by, cy, ax & by, ax & cy bx & ay, bx & cy cx & ay, cx & by 12 different ways. (or 3(1) + 3(1) + 3(2)) the total number of ways for 40 people to be matched with two different prizes is (40)(39) (permutation formula because order matters) so the prob ...

actually, it doesn't matter if the prizes are different or not. the # ways for the family to get a prize is (3 ch 1) + (3 ch 2) = 3+3 = 6 the total number of ways the prizes can be given out is (40 ch 2) = (40x39)/2 6/((40x39)/2) = 12/(40x39) same as before

Cristofer, thanks for replying. Well, the answer is somehow 19/130...

wow, i totally messed that up! sorry, but I see my mistake now. the numerator should be (3 ch 1)(37 ch 1) + (3 ch 2)(37 ch 0), which is 3x37 + 3 = 3x38 so (3x38)/((40x39)/2) = (6x38)/(40x39) = 19/130 sorry! i hope you can forgive me

Thanks a lot for helping!!! BTW a friend of mine came up with an easy solution. The calculation she used goes from the opposite - what's the probability that none of the 3 family members gets a prize: 37/40 x 36/39 = 111/130 Now substract this from the total number of probabilities: 1-111/130=19/130

Probability of winning atleast 1 of the raffles = 1 - (probability of winning nothing) Prob of winning nothing on the first attemp = 37/40 Prob of winning nothing on the second attemp = 36/39 Prob of winning nothing = P(a) and P(b) Prob of winning atleast 1 = 1- (37/40*36/39) = 19/130