I will assume the chips are not returned after being selected.
a)
You want the situation
GGGGN and its other 4 variations, where N stands for non-green
Prob(GGGGN) = (30/60)(29/59)(28/58)(27/57)(30/56)
= 135/4484
But there are C(5,4) of these
Prob(exactly 4 green) = 5(135/4484) = 675/4484
b) at least 2 red ---> exlude 0 red and 1 red
prob(zero red) = prob(NNNNN)
= (50/60)(49/59)(48/58)(47/57)(46/56)
= (50x49x48x47x46)/(60x59x58x57x56)
= 117/4484
prob(1 red) = 5x(50/60)(49/59)(48/58)(47/57)(10/56)
= .......
prob(at least 2 red)
= 1 - (prob(zero red) + prob(1 red) )
= ...
c) at most 2 blue
= prob(zero blue) + prob(1 blue) + prob(2 blue)
check my arithmetic and typing of the set-up.
I am only on my second cup of coffee.
Can someone help me answer this question, so I would know what to do for a question like this next time?
Thank you!
A box contains 10 red chips, 20 blue chips, and 30 green chips. If 5 chips are drawn from the box, find the probability of drawing
a.) exactly 4 green
b.) at least 2 red
c.) at most 2 blue
1 answer
1 answer