Question 241278: You order twelve burritos to go from a Mexican restaurant, five with hot peppers and seven without. However, the restaurant forgot to label them. If you pick three burritos at random, find the probability that at least two have hot peppers. (Round the answer to three decimal places.)
My mother told me 0.36 but my teacher said it was wrong. Could you please help me.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you have 12 burritos to pick from.
5 are hot.
seven are not.
you pick 3 at random.
probability that at least 2 have hot peppers would be:
???
the probability of getting exactly 3 hot ones would be:
(5*4*3)/(12*11*10) = 60/1320
the probability of getting exactly 2 hot ones would be:
(5*4*7)/(12*11*10) * 3 = 120/1320 * 3 = 420/1320
Any other possible picks would be invalid because they would involve less than 2 burritos.
the probability of getting exactly 3 hot burritos can happen in 1 way only.
HHH
the probability of getting exactly 2 hot burritos can happen in 3 ways only.
HHC
HCH
CHH
that's why we multiplied by 3 to get 420/1320.
since the probability of exactly 3 or exactly 2 is additive, then we have 60/1320 + 420/1320 = 480/1320 which is equal to .36363636364 which, if you round to 3 decimal places, equals .364.
If we did this correctly, then the total probability should be 1.
p(0H) = 1 * (7*6*5)/1320 = 210/1320 = 210/1320
p(1H) = 3 * (5*7*6)/1320 = 210/1320 = 630/1320
p(2H) = 3 * (5*4*7)/1320 = 140/1320 = 420/1320
p(3H0 = 1 * (5*4*3)/1320 = 60/1320 = 60/1320
total probability is (210+630+420+60) / 1320 = 1320 / 1320 = 1
total probability should be 1 and it is so your mother is right.
only problem I can find is in the rounding.
Answer should be .364 and not .36.
Other than that I think the answer that your mother gave you is correct.
|
|
|
