Probability Problem

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adi: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Mon Jan 07, 2008 12:55 pm; Location: Frisco, Texas

by adi » Fri Mar 14, 2008 7:47 pm

There are 15 slate rocks, 20 pumice rocks, and 10 granite rocks randomly distributed in a certain field. If 2 rocks are to be chosen at random and without replacement, what is the probability that both rocks will be slate rocks?

A) 1/3
B) 7/22
C) 1/9
D) 7/66
E) 2/45

Answer should be D.

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Source: Beat The GMAT — Problem Solving | Fri Mar 14, 2008 7:47 pm

musicdaemon: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Wed Feb 13, 2008 10:34 am; Location: India; Thanked: 3 times

by musicdaemon » Fri Mar 14, 2008 9:03 pm

This is a fairly basic one, anyway here is the solution

We Know,
Probability for the Event n(P) = n(E)/n(S)

Where n(E) = Number of events that the two rocks selected are Slate stone
n(S) = Sample Size ( number of ways to pick any 2 stones)

Here,
n(E) = 15C2 = 15!/(2!*13!) = 15*7
n(S) = 45C2 = 45!/(2!*43!) = 45*22
Therefore,
n(P) = n(E)/n(S) = (15*7)/( 45*22) = 7/66 ………………. Ans

Let the Game begin

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yev23: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Tue Mar 04, 2008 11:11 amFollowed by: 1 member; Location: Boston, MA

by yev23 » Mon Mar 17, 2008 4:42 pm

Another way of thinking about is:

1. Probability of getting first slate rock: 15/45 = 1/3
2. Probability of getting 2nd slate rock: 14/44 = 7/22

Thus: (1/3)*(7/22) = 7/66

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