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A box of 30 flashbulbs contains 3 defective . A random sample of 2 is selected and tested.
Let X be the random variable associated with the number of defective bulbs in the sample.
(A) Find the probability distribution of X.
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As this problem is worded, it tells me that it assigned to a person
who has some necessary prerequisites and understand simple separated cases.
So I will not go in detailed explanations.
The random variable X may have one of three values: X= 0, 1, 2.
P(X=0) is the probability that a random sample of 2 flashbulbs contains 0 defective.
The probability for it is P(X=0) = = = = 0.806896552 (rounded).
P(X=1) is the probability that a random sample of 2 flashbulbs contains 1 defective.
The probability for it is P(X=1) = = = = 0.186206897 (rounded).
P(X=2) is the complement of P(X=0) + P(X=1) to 1, so we calculate it this way
P(X=2) = 1 - 0.806896552 - 0.186206897 = 0.006896551.
We can check the value P(X=2) this way P(X=2) = = = . 117+27+1 = 145. ! correct !
Solved.