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Updated on: 10 Jul 2019, 23:15
Originally posted by rohan2345 on 20 May 2017, 06:39.
Last edited by Sajjad1994 on 10 Jul 2019, 23:15, edited 1 time in total.
Last edited by Sajjad1994 on 10 Jul 2019, 23:15, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
63% (01:23) correct
37%
(01:23)
wrong
based on 136
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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
(A) 1/12
(B) 1/8
(C) 1/6
(D) 1/4
(E) 1/3
Source: Nova GMAT
Difficulty Level: 600
Difficulty Level: 600
20 May 2017, 06:46
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rohan2345
\(P(sum = 3) = P(0, 3) + P(1, 2) = 2*\frac{1}{4}*\frac{1}{3} + 2*\frac{1}{4}*\frac{1}{3} = \frac{1}{3}\). 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).
Answer: E.
20 May 2017, 11:24
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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
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
