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GMAT Club Forum INDEX Problem Solving (PS)

A certain bag of gemstones is composed of two-thirds : Problem Solving (PS)

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BG

May 29, 2006

00:00ABCDE

DIFFICULTY: 65% (hard) QUESTION STATS: based on 338 sessions

64% (02:50) correct

36% (02:44) wrong

A certain bag of gemstones is composed of two-thirds diamonds and one-third rubies. If the PROBABILITY of randomly SELECTING two diamonds from the bag, without replacement, is 5/12, what is the probability of selecting two rubies from the bag, without replacement?

(A) 5/36

(B) 5/24

(C) 1/12

(D) 1/6

(E) 1/4

Spoiler: OA

Kudos

8 kudos, 23 bookmarks

Most Helpful Expert Reply

VeritasKarishma

EXPERT'S

POST

Jan 9, 2014

b00gigi wrote:

Can someone explain the

2/3 * (2R-1)/(3R-1)

part?

Say, a bag has 6 diamonds and 3 rubies. What is the probability of selecting 2 diamonds one after the other without replacement?

Probability of selecting one diamond = 6/9

Probability of selecting yet another diamond after selecting one = 5/8 (no of diamonds has gone down by 1 and total no. of diamonds has gone down by 1 too)

Total probability = (6/9)*(5/8)

Here, we assume that no of rubies is R and no of diamonds is 2R (since no of diamonds is twice the no of rubies)

Probability of selecting two diamonds without replacement = (2R/3R) * (2R - 1)/(3R - 1) = 5/12

Either cross multiply to get the value of R or try to plug in some values to see where you get a multiple of 12 in the denominator.

Once you get the value of R as 3, you know the number of diamonds is 6.

Probability of picking two rubies one after the other without replacement = (3/9) *(2/8) = 1/12