Bunuel wrote:

Three of 12 remote controls in a box are defective. The remote controls are tested one at a time and not replaced in the box after testing. What is the probability that neither of the first two remote controls tested is defective?

(A) 1/6

(B) 2/9

(C) 6/11

(D) 9/16

(E) 3/4

Since there are 9 non-defective remotes out of 12 in the box, the probability that the first remote selected is not defective is 9/12, or 3/4. We don’t put the first selected remote control back in the box. Thus, there are now 11 remotes in the box, and 8 of them are not defective. Thus, the probability that the second remote control is not defective is 8/11.

Thus, the probability that the first two remote controls are not defective is 9/12 x 8/11 = 3/4 x 8/11 = 24/44 = 6/11.

Answer: C

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Scott Woodbury-Stewart

Founder and CEO

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