Answer:
[tex]\begin{array}{cc}X&P(X)\\0&0\\1&0.3333\\2&0.6667\end{array}[/tex]
E(X)= 1.667
Step-by-step explanation:
The question asks for the probability distribution and the expected value of X.
The possible values for the number of defective cameras (X) are 0, 1 or 2.
The probability distribution for X is:
[tex]P(X=0)=0[/tex]
Since there is only one camera in the box that is not defective, it is impossible for no camera to be defective when picking 2.
[tex]P(X=1)=\frac{1}{6}*\frac{5}{5}+\frac{5}{6}*\frac{1}{5}\\ P(X=1)=0.3333[/tex]
[tex]P(X=2)=\frac{5}{6}*\frac{4}{5}\\ P(X=2) =0.6667[/tex]
The probability distribution is:
[tex]\begin{array}{cc}X&P(X)\\0&0\\1&0.3333\\2&0.6667\end{array}[/tex]
The expected value of X is:
[tex]E(X) = 0.3333*1+0.6667*2\\E(X) = 1.667[/tex]
The expected value is 1.667 defective cameras.
