Question 1201802
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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.



<pre>
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) = {{{(27/30)*(26/29)}}} = {{{702/870}}} = {{{117/145}}} = 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) = {{{2*(27/30)*(3/29)}}} = {{{162/870}}} = {{{27/145}}} = 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) = {{{(3/30)*(2/29)}}} = {{{2/290}}} = {{{1/145}}}.  117+27+1 = 145.  ! correct !
</pre>

Solved.