PrepLattice is an independent test-preparation service and is not affiliated with or endorsed by GMAC, the organization that administers the GMAT. GMAT and GMAT Focus are trademarks of GMAC, used here only to name the exam this question is designed to prepare you for.
A bin contains 8 identical-looking widgets, of which exactly 3 are defective and 5 are good. An inspector draws the widgets one at a time at random without replacement. What is the probability that the second defective widget is drawn on exactly the 4th draw?
Five fresh questions every day, your progress tracked, every miss explained. Free with an account.
Correct answer
B
The second defective falls on the 4th draw exactly when the first three draws hold exactly one defective and the 4th draw is defective. The chance of exactly one defective among the first three is C(3,1) × C(5,2) ÷ C(8,3) = (3 × 10) ÷ 56 = 30/56 = 15/28. After those three draws, 1 defective and 2 good are gone, leaving 2 defective and 3 good among 5, so the 4th draw is defective with probability 2/5. Multiply the two stages: (15/28) × (2/5) = 30/140 = 3/14.
So the answer is B.
Pick the theme that suits how you work. You can change it anytime.