It is currently 19 Apr 2018, 08:33

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Moderator
Joined: 29 Jan 2015
Posts: 720
Location: India
WE: General Management (Non-Profit and Government)
A bowl contains one marble labeled 0, one marble labeled 1 [#permalink]

### Show Tags

20 May 2017, 06:39
1
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

64% (00:57) correct 36% (00:43) wrong based on 81 sessions

### HideShow timer Statistics

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.
[Reveal] Spoiler: OA

_________________

If you liked my post, kindly give me a Kudos. Thanks.

Math Expert
Joined: 02 Sep 2009
Posts: 44566
Re: A bowl contains one marble labeled 0, one marble labeled 1 [#permalink]

### Show Tags

20 May 2017, 06:46
1
KUDOS
Expert's post
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).

_________________
DS Forum Moderator
Joined: 22 Aug 2013
Posts: 1010
Location: India
Re: A bowl contains one marble labeled 0, one marble labeled 1 [#permalink]

### Show Tags

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

Intern
Joined: 10 Sep 2017
Posts: 6
Re: A bowl contains one marble labeled 0, one marble labeled 1 [#permalink]

### Show Tags

09 Apr 2018, 18:31
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: 44566
Re: A bowl contains one marble labeled 0, one marble labeled 1 [#permalink]

### Show Tags

09 Apr 2018, 21:32
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?

_________________
Re: A bowl contains one marble labeled 0, one marble labeled 1   [#permalink] 09 Apr 2018, 21:32
Display posts from previous: Sort by