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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

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.

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

This post received 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

\(= \frac{25}{216} * 3\)

\(= \frac{25}{72}\)

Hence, Answer is C _________________

"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

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,

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:

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,