SOLUTION: In a bag of 30 chocolates, 16 have vanilla filling and 14 have strawberry filling. Six chocolates are randomly chosen from the bag. a) What is the probability of obtaining exact

Question 1089049: In a bag of 30 chocolates, 16 have vanilla filling and 14 have strawberry filling. Six
chocolates are randomly chosen from the bag.
a) What is the probability of obtaining exactly 2 chocolates with strawberry filling?
b) What is the expected number of vanilla filling chocolates?
i need help right now plzzzzzzz.
Answer by natolino_2017(77) (Show Source): You can put this solution on YOUR website!
Let x number of Chocolates extracted with strawberry filling.
a) P(x=2) = (14C2)((30-14)C(10-2))/(30C10) = (14C2)(16C8)/(30C10) = (91)(12,870)/(30,045,015) = 26/667 = 0.03898
so the answer is close to the 4%.
Let y number of number of chocolates extracted with vainilla filling
we need the Probabilities of P(y=a) a=1,2,...10 (zero is optional)
b) P(y=1) =(16C1)((30-16)C(10-1)/(30C10) = (16C1)(14C9)/(30C10 = (16)(2,002)/(30,045,015) = 32/30,015.
P(y=2) =(16C2)((30-16)C(10-2)/(30C10) = (16C2)(14C8)/(30C10 = (120)(3003)/(30,045,015) = 8/667
repeat the procedure to obtain the rest:
P(y=3) = 64/10,005
P(y=4) = 364/2,001
P(y=5) = 2,912/10,005
P(y=6) = 8,008/30,015
P(y=7) = 832/6,003
P(y=8) = 26/667 (which is the same as question a))
P(y=9) = 32/6,003
P(y=10) = 8/30,015
Now Expect Value of y is:
E(y) = Sum(a*P(y=a)) from a=0 to 10 (that's why the P(y=0) is optional).
= 16/3 = 5.33333333333333333333
So the expected value is closer to 5 chocolates
***Observation: if the number of strawberry and vanilla would be the same, the expected value would be exactly = 10/2 = 5. ***
@natolino_

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