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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