It is currently 19 Jan 2018, 01:55

### 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

# If a fair 6-sided die is rolled three times, what is the probability t

Author Message
TAGS:

### Hide Tags

Manager
Joined: 26 Nov 2014
Posts: 106

Kudos [?]: 115 [0], given: 8

If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

### Show Tags

04 Nov 2015, 04:15
8
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

57% (00:38) correct 43% (00:26) wrong based on 259 sessions

### HideShow timer Statistics

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6
[Reveal] Spoiler: OA

_________________

Consider Kudos for my post, if it is helpful.
TIA

Kudos [?]: 115 [0], given: 8

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7866

Kudos [?]: 18473 [2], given: 237

Location: Pune, India
Re: If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

### Show Tags

04 Nov 2015, 04:39
2
KUDOS
Expert's post
7
This post was
BOOKMARKED
shyind wrote:
If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6

Total ways in which a 6-sided die can be rolled three times = 6*6*6 = 216

To get exactly one 3, there are three ways:
A 3 on the first roll and non 3 on other two rolls. This can be done in 1*5*5 = 25 ways.
The 3 could be on the second or third roll too. So total favorable cases = 25*3 = 75

Required Probability = 75/216 = 25/72

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 18473 [2], given: 237

Intern
Joined: 24 Sep 2015
Posts: 4

Kudos [?]: 8 [0], given: 2

Re: If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

### Show Tags

04 Nov 2015, 20:22
shyind wrote:
If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6

Probability if getting 3 on a die is: P(3)=1/6
Probability of not getting 3 on a die is nP(3)=5/6

Let the three dies be denoted by P1,P2 and P3

Probability of getting 3 on the first die and not getting 3 on the other two dies is given by:-
=P1(3)*nP(3)*nP(3)
=1/6*5/6*5/6
=25/216

As there are 3 dies,so similarly for the rest of the two dies probability will be 25/216

Summing up all the 3 cases we get:
=3*25/216
=25/72

Kudos [?]: 8 [0], given: 2

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4692

Kudos [?]: 3317 [0], given: 0

GPA: 3.82
Re: If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

### Show Tags

05 Nov 2015, 00:22
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6

Let O be represented as 3 being rolled, and X be represented as no 3 being rolled.

Then the case of exactly one 3 being rolled can be represented as (O, X, X), (X, O, X), (X, X, O).
The probability of (O, X, X) is (1/6)*(5/6)*(5/6), that of (X, O, X) is (5/6)*(1/6)*(5/6) and that of (X, X, O) is (5/6)*(5/6)*(1/6).
So the probability is (1/6)*(5/6)*(5/6) + (5/6)*(1/6)*(5/6) + (5/6)*(5/6)*(1/6) = 3*(1/6)*(5/6)*(5/6) = 25/72.

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself

Kudos [?]: 3317 [0], given: 0

Verbal Forum Moderator
Joined: 19 Mar 2014
Posts: 979

Kudos [?]: 273 [2], given: 199

Location: India
Concentration: Finance, Entrepreneurship
GPA: 3.5
Re: If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

### Show Tags

08 Jul 2017, 11:50
2
KUDOS
shyind wrote:
If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

A. 25/216

B. 50/216

C. 25/72

D. 25/36

E. 5/6

Probability of getting 3 $$= \frac{1}{6}$$

Probability of not getting 3 $$= \frac{5}{6}$$

Probability of getting three on the first roll $$= \frac{1}{6} * \frac{5}{6} * \frac{5}{6}$$

Probability of getting three on the first roll $$= \frac{25}{216}$$

As we can get 3 on either second or the third time we will multiply the above probability by 3

$$= \frac{25}{216} * 3$$

$$= \frac{25}{72}$$

_________________

"Nothing in this world can take the place of persistence. Talent will not: nothing is more common than unsuccessful men with talent. Genius will not; unrewarded genius is almost a proverb. Education will not: the world is full of educated derelicts. Persistence and determination alone are omnipotent."

Worried About IDIOMS? Here is a Daily Practice List: https://gmatclub.com/forum/idiom-s-ydmuley-s-daily-practice-list-250731.html#p1937393

Best AWA Template: https://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html#p470475

Kudos [?]: 273 [2], given: 199

Veritas Prep GMAT Instructor
Joined: 01 Jul 2017
Posts: 26

Kudos [?]: 16 [0], given: 3

Re: If a fair 6-sided die is rolled three times, what is the probability t [#permalink]

### Show Tags

06 Jan 2018, 16:32
Here is the full answer, from the ground up. To understand the mathematics behind this question we need to understand three basic probability rules:

(1) A probability is calculated by dividing the number of options when the event occurs by the total number of possibilities.

(2) When calculating the probability of multiple events that must ALL occur to meet a certain condition, we must multiply the separate probabilities of each event together. Thus, the probability of all the events $$A$$, $$B$$, and $$C$$ occurring would be: $$P_{combined} = (P_A)*(P_B)*(P_C)$$

(3) When calculating the probability of a series of mutually-exclusive events, where ANY event occurring would be sufficient, we must add the separate probabilities of each event together. Thus, the probability of any of the events $$A$$, $$B$$, and $$C$$ occurring would be: $$P_{combined} = (P_A)+(P_B)+(P_C)$$

Let's leverage these three basic rules to solve this problem.

First, the probability of rolling a single value (in this case 3) on a fair 6-sided die would be one out of six (or $$\frac{1}{6}$$).

The probability of NOT rolling that number would be five out of six (or $$\frac{5}{6}$$).

So, if you need to roll a single value on one die, while simultaneously not rolling that value on the other two dice, then we invoke rule #2 from above (i.e., we multiply the probabilities.) Thus,

$$P=(\frac{1}{6})*(\frac{5}{6})*(\frac{5}{6})=\frac{25}{216}$$

Notice that this is answer choice A. However, this is a deliberate trap answer. With probabilities, the order always matters. (I like to call this "100% Chance of Order" in my classes.) Thus, if we don't care which of the dice roll the "3", then we need to look at each of the possible orders that could work. There are three possibilities:

$$3, N, N$$
$$N, 3, N$$
$$N, N, 3$$

The probability of each of these orders is the same ($$\frac{25}{216}$$), but each is separate and independent. We now need to invoke rule #3 from above and add the separate probabilities together:

$$P_{Total} = \frac{25}{216} + \frac{25}{216} + \frac{25}{216} = \frac{3*25}{216}$$

Recognizing common factors in the top and bottom of this fraction allow us to get to the answer without any messy math. $$216 = 6*6*6 = 2*3*6*6$$. Thus,

$$\frac{3*25}{216}=\frac{3*25}{2*3*6*6}=\frac{25}{2*6*6}=\frac{25}{72}$$

_________________

Aaron J. Pond
Veritas Prep Elite-Level Instructor

Hit "+1 Kudos" if my post helped you understand the GMAT better.
Look me up at https://www.veritasprep.com/gmat/aaron-pond/ if you want to learn more GMAT Jujitsu.

Kudos [?]: 16 [0], given: 3

Re: If a fair 6-sided die is rolled three times, what is the probability t   [#permalink] 06 Jan 2018, 16:32
Display posts from previous: Sort by