The Expected value of XX is 1.00.
Given that a box contains 8 cameras and that 4 of them are defective and 2 cameras is selected at random with replacement.
The probability distribution of the hypergeometric is as follows:
Where x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.
The probability distribution for X is obtained as below:
From the given information, let X be a random variable, that denotes the number of defective cameras following hypergeometric distribution.
Here, M = 4, n=2 and N=8
The probability distribution of X is obtained below:
The probability distribution of X is,
The probability distribution of X when X=0 is
![\begin{aligned}P(X=0)&=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}8-4\\ 2-0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}4\\ 2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-0)!0!}\right)\times \left(\frac{4!}{(4-2)!2!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DP%28X%3D0%29%26%3D%5Cfrac%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%200%5Cend%7Barray%7D%5Cright%29%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D8-4%5C%5C%202-0%5Cend%7Barray%7D%5Cright%29%7D%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D%208%5C%5C%202%5Cend%7Barray%7D%5Cright%29%7D%5C%5C%20%26%3D%5Cfrac%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%200%5Cend%7Barray%7D%5Cright%29%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%202%5Cend%7Barray%7D%5Cright%29%7D%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D%208%5C%5C%202%5Cend%7Barray%7D%5Cright%29%7D%5C%5C%20%26%3D%5Cfrac%7B%5Cleft%5B%5Cleft%28%5Cfrac%7B4%21%7D%7B%284-0%29%210%21%7D%5Cright%29%5Ctimes%20%5Cleft%28%5Cfrac%7B4%21%7D%7B%284-2%29%212%21%7D%5Cright%29%5Cright%5D%7D%7B%5Cleft%28%5Cfrac%7B8%21%7D%7B%288-2%29%212%21%7D%5Cright%29%7D%5C%5C%20%26%3D0.21%5Cend)
The probability distribution of X when X=1 is
![\begin{aligned}P(X=1)&=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}8-4\\ 2-1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-1)!1!}\right)\times \left(\frac{4!}{(4-1)!1!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.57\end](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DP%28X%3D1%29%26%3D%5Cfrac%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%201%5Cend%7Barray%7D%5Cright%29%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D8-4%5C%5C%202-1%5Cend%7Barray%7D%5Cright%29%7D%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D%208%5C%5C%202%5Cend%7Barray%7D%5Cright%29%7D%5C%5C%20%26%3D%5Cfrac%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%201%5Cend%7Barray%7D%5Cright%29%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%201%5Cend%7Barray%7D%5Cright%29%7D%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D%208%5C%5C%202%5Cend%7Barray%7D%5Cright%29%7D%5C%5C%20%26%3D%5Cfrac%7B%5Cleft%5B%5Cleft%28%5Cfrac%7B4%21%7D%7B%284-1%29%211%21%7D%5Cright%29%5Ctimes%20%5Cleft%28%5Cfrac%7B4%21%7D%7B%284-1%29%211%21%7D%5Cright%29%5Cright%5D%7D%7B%5Cleft%28%5Cfrac%7B8%21%7D%7B%288-2%29%212%21%7D%5Cright%29%7D%5C%5C%20%26%3D0.57%5Cend)
The probability distribution of X when X=2 is
![\begin{aligned}P(X=2)&=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}8-4\\ 2-2\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left(\begin{array}{l}4\\ 2\end{array}\right)\left(\begin{array}{l}4\\ 0\end{array}\right)}{\left(\begin{array}{l} 8\\ 2\end{array}\right)}\\ &=\frac{\left[\left(\frac{4!}{(4-2)!2!}\right)\times \left(\frac{4!}{(4-0)!0!}\right)\right]}{\left(\frac{8!}{(8-2)!2!}\right)}\\ &=0.21\end](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DP%28X%3D2%29%26%3D%5Cfrac%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%202%5Cend%7Barray%7D%5Cright%29%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D8-4%5C%5C%202-2%5Cend%7Barray%7D%5Cright%29%7D%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D%208%5C%5C%202%5Cend%7Barray%7D%5Cright%29%7D%5C%5C%20%26%3D%5Cfrac%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%202%5Cend%7Barray%7D%5Cright%29%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D4%5C%5C%200%5Cend%7Barray%7D%5Cright%29%7D%7B%5Cleft%28%5Cbegin%7Barray%7D%7Bl%7D%208%5C%5C%202%5Cend%7Barray%7D%5Cright%29%7D%5C%5C%20%26%3D%5Cfrac%7B%5Cleft%5B%5Cleft%28%5Cfrac%7B4%21%7D%7B%284-2%29%212%21%7D%5Cright%29%5Ctimes%20%5Cleft%28%5Cfrac%7B4%21%7D%7B%284-0%29%210%21%7D%5Cright%29%5Cright%5D%7D%7B%5Cleft%28%5Cfrac%7B8%21%7D%7B%288-2%29%212%21%7D%5Cright%29%7D%5C%5C%20%26%3D0.21%5Cend)
Use E(X)=∑xP(x) to find the expected values of a random variable X.
The expected values of a random variable X is obtained as shown below:
The expected value of X is,
E(X)=∑xP(x-X)
E(X)=[(0×0.21)+(1×0.57)+(2×0.21)]
E(X)=[0+0.57+0.42]
E(X)=0.99≈1
Hence, the binomial probability distribution of XX when X=0 is 0.21, when X=1 is 0.57 and when X=2 is 0.21 and the expected value of XX is 1.00.
Learn about Binomial probability distribution from here brainly.com/question/10559687
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