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

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

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Updated on: 10 Jul 2019, 23:15
2
00:00

Difficulty:

45% (medium)

Question Stats:

62% (01:27) correct 38% (01:22) wrong based on 97 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
Difficulty Level: 600

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Originally posted by rohan2345 on 20 May 2017, 06:39.
Last edited by SajjadAhmad on 10 Jul 2019, 23:15, edited 1 time in total.
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Re: A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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20 May 2017, 06:46
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, 11:24
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

Hence answer is E
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Re: A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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18 Feb 2019, 11:48
Ways to select 2 numbers that equal 3 = 2 ways
Ways to arrange them = 2 ways (1+2, 2+1 and 0+3, 3+0)
Total choices = 4C1*3C1

2*2 / 4*3 = 4/12 = 1/3
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Re: A bowl contains one marble labeled 0, one marble labeled 1  [#permalink]

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18 Feb 2019, 14:09
No matter what marble you pick first, you can always make a sum of 3 if you pick the right marble second. And when you pick the second marble, you are picking from 3 marbles, exactly one of which gives us the right sum, so the answer is 1/3.
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Re: A bowl contains one marble labeled 0, one marble labeled 1   [#permalink] 18 Feb 2019, 14:09
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# A bowl contains one marble labeled 0, one marble labeled 1

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