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Suppose that a box contains 6 cameras and that 4 of them are defective. A sample of 2 cameras is selected at random, with replacement. Define the random variable
as the number of defective cameras in the sample.

Write the probability distribution for X.


What is the expected value of X
?



Answer :

TiltedLunar

1. Probability distribution:

  P(X = 0) = (2/6) * (2/6) = 4/36 = 1/9

  P(X = 1) = (4/6) * (2/6) + (2/6) * (4/6) = 16/36 = 4/9

  P(X = 2) = (4/6) * (4/6) = 16/36 = 4/9

2. Expected value calculation:

  E(X) = 0 * P(X=0) + 1 * P(X=1) + 2 * P(X=2)

       = 0 * (1/9) + 1 * (4/9) + 2 * (4/9)

       = 0 + 4/9 + 8/9

       = 12/9

       = 4/3

Therefore, the expected value of X is 4/3 or approximately 1.33 defective cameras in the sample.

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