Please explain the problem to me step by step
1 answer:
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Step-by-step explanation:
From the given gummi bear, the chance that Adam is selected for any draw is 1/3 as well as the chance he is not selected at any draw is 2/3.
a). The probability of Adam getting exactly three gummi bears = P(Adam gets selected at 3 draws and not selected at the remaining 9 draws)
=
Now, the 3 draws where Adam gets selected can be any 3 out of 12 draws in = 220
Thus, probability of Adam getting three gummi bears =
= 0.21186
b). Probability that Adam will get the three gummi bears given each person will received at the most 1 gummi bear
= P(of the remaining 9 draws after assigning one gummi bear to each one, Adam gets selected at 2 draws and not selected at 7 draws) =
= 0.23411
c). Let X = Number of the gummi bears which Adam will get. Then, X = number of draws out of 12 draws Adam gets selected and X ~ B(12, 1/3). So, Adam will get gummi bears= mean of B(12, 1/3) = 12 x (1/3) = 4
d). Let Y = Number of the gummi bears that Adam will get, given each person will received at the most 1 gummi bear Then, Y = number of draws Adam gets selected in the remaining 9 draws and Y ~ B(9, 1/3). So, the expected number of bears that Adam gets given each person received at least 1 gummi bear
= 1 + mean of B(9, 1/3) = 4
e). When every one gets atleast one gummi bear, the new sample size will be 9 and so we can say that there is a reduction in variance.