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At a certain car dealership, the probability of selecting a car of red [#permalink]
Bunuel wrote:
At a certain car dealership, the probability of selecting a car of red color by customer is 1/6. The probability of selecting a car of X model is 5/13. Which of the following could not be the probability of selecting a car, which is either of red color or X model or both?

(A) 3/8
(B) 5/11
(C) 5/13
(D) 7/13
(E) 8/15



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Probabilities as follows:

Red/X + Red/no X = 1/6

Red/X + not Red/X = 5/13



Question asks for potential probability of:

Red/X + Red/no X + not Red/X



This equals 5/13 + Red/no X

Examining A:

Can 5/13 + Red/no X = 3/8 ?


5/13 = 40/104. 3/8 = 39/104

A can't be the answer since it would require Red/no X to be negative.

Likewise one could use 1/6 + not Red/X and that would require Red/no X to be negative, also not possible.

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At a certain car dealership, the probability of selecting a car of red [#permalink]
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