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# A bowl contains one marble labeled 0, one marble labeled 1

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General GMAT Forum Moderator
Joined: 29 Jan 2015
Posts: 1125
Location: India
WE: General Management (Non-Profit and Government)
A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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20 May 2017, 05:39
1
00:00

Difficulty:

35% (medium)

Question Stats:

67% (01:31) correct 33% (01:27) wrong based on 92 sessions

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A bowl contains one marble labeled 0, one marble labeled 1, one marble labeled 2, and one marble labeled 3. The bowl contains no other objects. If two marbles are drawn randomly without replacement, what is the probability that they will add up to 3?

(A) 1/12
(B) 1/8
(C) 1/6
(D) 1/4
(E) 1/3

Source: Nova GMAT.

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Math Expert
Joined: 02 Sep 2009
Posts: 52905
Re: A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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20 May 2017, 05:46
1
rohan2345 wrote:
A bowl contains one marble labeled 0, one marble labeled 1, one marble labeled 2, and one marble labeled 3. The bowl contains no other objects. If two marbles are drawn randomly without replacement, what is the probability that they will add up to 3?

(A) 1/12
(B) 1/8
(C) 1/6
(D) 1/4
(E) 1/3

Source: Nova GMAT.

$$P(sum = 3) = P(0, 3) + P(1, 2) = 2*\frac{1}{4}*\frac{1}{3} + 2*\frac{1}{4}*\frac{1}{3} = \frac{1}{3}$$. We multiply by 2 in each case because (0, 3) can occur in two ways first 0, then 3 OR first 3, then 0. The same for (1, 2).

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Re: A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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20 May 2017, 10:24
Probability of an event = (Number of favourable outcomes)/(Total Outcomes)

Here total outcomes = total ways of selecting (or drawing) 2 marbles out of 4
=4C2 = 4!/2!2! = 6.

Favoarble outcomes = total ways where the numbers on two marbles add up to 3
= 2 (either numbers 0 & 3 OR numbers 1 & 2)

So required probability = 2/6 = 1/3

Intern
Joined: 10 Sep 2017
Posts: 5
Re: A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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09 Apr 2018, 17:31
Bunuel wrote:
rohan2345 wrote:
A bowl contains one marble labeled 0, one marble labeled 1, one marble labeled 2, and one marble labeled 3. The bowl contains no other objects. If two marbles are drawn randomly without replacement, what is the probability that they will add up to 3?

(A) 1/12
(B) 1/8
(C) 1/6
(D) 1/4
(E) 1/3

Source: Nova GMAT.

$$P(sum = 3) = P(0, 3) + P(1, 2) = 2*\frac{1}{4}*\frac{1}{3} + 2*\frac{1}{4}*\frac{1}{3} = \frac{1}{3}$$. We multiply by 2 in each case because (0, 3) can occur in two ways first 0, then 3 OR first 3, then 0. The same for (1, 2).

2*1/4*1/3 = 1/6? How do you get 1/6 + 1/6 = 1/3?
Math Expert
Joined: 02 Sep 2009
Posts: 52905
Re: A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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09 Apr 2018, 20:32
cman2010 wrote:
Bunuel wrote:
rohan2345 wrote:
A bowl contains one marble labeled 0, one marble labeled 1, one marble labeled 2, and one marble labeled 3. The bowl contains no other objects. If two marbles are drawn randomly without replacement, what is the probability that they will add up to 3?

(A) 1/12
(B) 1/8
(C) 1/6
(D) 1/4
(E) 1/3

Source: Nova GMAT.

$$P(sum = 3) = P(0, 3) + P(1, 2) = 2*\frac{1}{4}*\frac{1}{3} + 2*\frac{1}{4}*\frac{1}{3} = \frac{1}{3}$$. We multiply by 2 in each case because (0, 3) can occur in two ways first 0, then 3 OR first 3, then 0. The same for (1, 2).

2*1/4*1/3 = 1/6? How do you get 1/6 + 1/6 = 1/3?

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Re: A bowl contains one marble labeled 0, one marble labeled 1   [#permalink] 09 Apr 2018, 20:32
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