GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 May 2019, 16:20

### 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

# Three fair six-sided dice are rolled. What is the probability that the

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55188
Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

### Show Tags

23 Apr 2019, 13:15
00:00

Difficulty:

85% (hard)

Question Stats:

38% (01:51) correct 62% (03:02) wrong based on 29 sessions

### HideShow timer Statistics

Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

(A) 1/6
(B) 13/72
(C) 7/36
(D) 5/24
(E) 2/9

_________________
Senior Manager
Joined: 19 Oct 2018
Posts: 250
Location: India
Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

### Show Tags

24 Apr 2019, 00:29
2
Total number of cases possible when 3 Dice roll= 6^3=216
Now we have to find the number of cases when values shown on two of the dice sum to the value shown on the remaining die-
1. When a dice shows 6, other 2 dices must show either (5,1) or (4,2) or (3,3)
Total number of cases possible in this scenario= 3!+3!+3!/2!=15
2. When a dice shows 5, other 2 dices must show either (4,1) or (3,2)
Total number of cases possible in this scenario= 3!+3!=12
3. When a dice shows 4, other 2 dices must show either (3,1) or (2,2)
Total number of cases possible in this scenario= 3!+3!/2!=9
4. When a dice shows 3, other 2 dices must show (2,1)
Total number of cases possible in this scenario= 3!=6
5. When a dice shows 2, other 2 dices must show (1,1)
Total number of cases possible in this scenario= 3!/2!=3
Total cases possible= 15+12+9+6+3=45

Prbability= 45/216=5/24
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6160
Location: United States (CA)
Re: Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

### Show Tags

25 Apr 2019, 19:18
Bunuel wrote:
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

(A) 1/6
(B) 13/72
(C) 7/36
(D) 5/24
(E) 2/9

There are 6^3 = 216 outcomes when three dices are rolled. If the sum of two of the dice is equal to the value of the remaining die. It could be:

1-1-2, 1-2-3, 1-3-4, 1-4-5, 1-5-6, 2-2-4, 2-3-5, 2-4-6, 3-3-6

For the combos above, each has 3!/2! = 3 outcomes (for example, 1-1-2 can also be 1-2-1 and 2-1-1). So the bold combos have a total 3 x 3 = 9 outcomes.

For the italic combos above, each has 3! = 6 outcomes (for example, 1-2-3 can also be 1-3-2, 2-1-3, 2-3-1. 3-1-2 and 3-2-1). So the italic combos have a total 6 x 6 = 36 outcomes.

Therefore, the probability that the values shown on two of the dice sum to the value shown on the remaining die is (9 + 36)/216 = 45/216 = 5/24.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager
Joined: 28 Jan 2019
Posts: 86
Location: India
Re: Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

### Show Tags

25 Apr 2019, 23:10
Bunuel wrote:
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

(A) 1/6
(B) 13/72
(C) 7/36
(D) 5/24
(E) 2/9

Let the die be A, B and C.
Given = A+B= C or A+C = B or B+C = A. ....... (I)

A dice has maximum number of 6. Therefore, the maximum sum of 2 numbers from the other two die would be 6.

Possibilities are:
1+1, 1+2, 1+3, 1+4, 1+5,
2+1, 2+2, 2+3, 2+4,
3+1, 3+2, 3+3,
4+1, 4+2,
5+1.
Totally, 15 possibilities.
Hence, probability = $$\frac{15}{6^3} = \frac{15}{216}$$

From (I), there are 3 such cases.
Hence, required probability = $$\frac{15}{216}*3 = \frac{45}{216} = \frac{5}{24}$$

_________________
"Luck is when preparation meets opportunity!"
CEO
Joined: 18 Aug 2017
Posts: 3473
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Three fair six-sided dice are rolled. What is the probability that the  [#permalink]

### Show Tags

27 Apr 2019, 01:23
Bunuel wrote:
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

(A) 1/6
(B) 13/72
(C) 7/36
(D) 5/24
(E) 2/9

total such pairs ;
(1,1,2) ( 1,2,3) , ( 1,3,4) , ( 1,4,5) , ( 1,5,6 )
(2,1,3) ( 2,2,4), ( 2,3,5) , ( 2,4,6)
(3,1,4) ( 3,2,5) , ( 3,3,6)
(4,1,5) ( 4,2,6)
(5,1,6)
15* 3; 45 pairs possible
total fair plays ; 6^ 3 ; 216
so P ; 45/216 ; 5/24
IMO D
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Re: Three fair six-sided dice are rolled. What is the probability that the   [#permalink] 27 Apr 2019, 01:23
Display posts from previous: Sort by