Marble Menace

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way2ashish: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Mon Sep 26, 2011 6:34 pm

by way2ashish » Sat Dec 29, 2012 2:17 am

Q) A bag contains some red and some blue marbles.How many red marbles are in the bag?
(1) If two marbles are selected from the bag, the probability of selecting one red and one blue marble is 1/4
(2) There are a total of 8 marbles in the bag.

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Source: Beat The GMAT — Data Sufficiency | Sat Dec 29, 2012 2:17 am

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by Anurag@Gurome » Sat Dec 29, 2012 10:57 am

way2ashish wrote:Q) A bag contains some red and some blue marbles.How many red marbles are in the bag?
(1) If two marbles are selected from the bag, the probability of selecting one red and one blue marble is 1/4
(2) There are a total of 8 marbles in the bag.

Statement 1: Say the total number of marbles in the bag is N and there are R red marbles. Hence, there are (N - R) blue marbles.

Number of ways to select two marbles from N marbles = NC2 = N(N - 1)/2
Number of ways to select one red marble from R marbles and one blue marbles from (N - R) marbles = (RC1)*((N - R)C1) = R*(N - R)

Hence, the probability of selecting one red and one blue marble = [R*(N - R)]/[N(N - 1)/2]
Hence, [R*(N - R)]/[N(N - 1)/2] = 1/4

As we have two unknowns N and R we cannot find the value of them from only one equation.

Statement 2: Clearly this statement is not sufficient

1 & 2 Together: N = 8
Now we have only one unknown R and we can solve the equation we obtained from statement 1 and find the value of R.

Sufficient

The correct answer is C.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Dec 29, 2012 11:24 pm

way2ashish wrote:Q) A bag contains some red and some blue marbles. How many red marbles are in the bag?
(1) If two marbles are selected from the bag, the probability of selecting one red and one blue marble is 1/4
(2) There are a total of 8 marbles in the bag.

Statement 2: The total number of marbles = 8.
Statement 1: P(one red and one blue) = 1/4.
The following cases satisfy both statements:

Case 1: B=7, R=1, for a total of 8 marbles.
P(BR) = 7/8 * 1/7 = 1/8.
P(RB) = 1/8 * 7/7 = 1/8.
Since either ordering (BR or RB) will yield one red marble and one blue marble, we add the fractions:
P(one red and one blue) = 1/8 + 1/8 = 1/4.
In this case, R=1.

Case 2: B=1, R=7, for a total of 8 marbles.
P(BR) =1/8 * 7/7 = 1/8.
P(RB) = 7/8 * 1/7 = 1/8.
Since either ordering (BR or RB) will yield one red marble and one blue marble, we add the fractions:
P(one red and one blue) = 1/8 + 1/8 = 1/4.
In the case, R=7.

Since the number of red marbles can be different values, the two statements combined are INSUFFICIENT.

The correct answer is E.

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mariofelixpasku: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Sun Sep 23, 2012 4:26 am; Thanked: 2 times

by mariofelixpasku » Thu Apr 04, 2013 12:14 pm

Mitch Hunt is right.

1.) R/(R+B) * B/(R+B-1) = 1/8 NS

2.) R/8 * B/7 = 1/8 NS

1.) + 2.) not sufficient

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J N: Senior | Next Rank: 100 Posts; Posts: 44; Joined: Tue Mar 12, 2013 2:27 pm; Thanked: 3 times

by J N » Fri Apr 05, 2013 9:36 pm

now redo the problem with probability = 4/7 and see what you get

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