gmatbusters wrote:
Only one of ten pens in a box is defective. Three pens are randomly drawn from the box, one at a time. What is the probability that the next pen randomly drawn is defective?
(A) 1/10
(B) 1/7
(C) 7/10
(D) 6/7
(E) 9/10
Source: Inspired from a Bunuel sir's question.
First three pens must be non-defective.
Probability to choose them:
9/10 * 8/9 * 7/8 ( we are not replacing the pens)
Fourth pen is defective:
1/7 is probability to choose it.
Total probability = 9/10*8/9*7/8*1/7 = 1/10
Hence Option
...
![FLO – Therapy at the Club – Pre-Single [iTunes Plus M4A]](http://is1-ssl.mzstatic.com/image/thumb/Music221/v4/a4/11/88/a41188ed-0f2e-e8d5-1163-d59e373cc58c/26UMGIM30046.rgb.jpg/208x208bb.webp)
.jpg)
