Question 972294: Given that there are 12 computers on a shelf at the store and 3 are defective; what is the probability of selecting A) four working computers, B) exactly two working computers and C) at least 1 working computer if a buyer selects four at random?
I tried to do 9C4/12C4 and believe I got part A. Would exactly 2 be 3C2 * 9C2 / 12C4? For part B. Part C seems to require null?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Yes to both.
Exactly two would be as you have written , and the probability = approx .22, which makes intuitive sense.
A check is that
(9/12)*(8/11)*(3/10)*(2/9) is one of six ways you can choose two working computers out of 4. The other five ways change the numerator for the denominator, but the products are the same, and they add up to the same number.
For C, yes, do the null. It is a lot easier.
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