Pam and Robin each roll a pair of fair, six-sided dice. What

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alanforde800Maximus: Master | Next Rank: 500 Posts; Posts: 187; Joined: Tue Sep 13, 2016 12:46 am

by alanforde800Maximus » Mon Dec 17, 2018 9:23 am

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Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?

A. 1/216

B. 1/36

C. 5/108

D. 11/216

E. 1/18

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Source: Beat The GMAT — Problem Solving | Mon Dec 17, 2018 9:23 am

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by [email protected] » Mon Dec 17, 2018 12:13 pm

Hi alanforde800Maximus,

This is a tougher probability question than average (and you likely will not see this exact situation on Test Day). This is meant to say that you shouldn't be too concerned about this prompt until you're picking up points in all the other 'gettable' areas first.

That having been said, the 'quirk' with this question is that you have to account for the probability of rolling two dice and getting the same number on both dice vs. getting two different numbers.

When rolling two dice, there are 36 possible outcomes:
-6 outcomes have the same number twice (1-1, 2-2, etc.)
-30 outcomes have two different numbers (1-4, 3-2, 5-1, etc.)

Thus, 6/36 = 1/6 of the outcomes are the same number twice
30/36 = 5/6 of the outcomes are two different numbers

IF Pam rolled 5-5, then Robin would have to also roll 5-5. The probability of that occurring would be: (1/6)(1/6) = 1/36.

IF Pam rolled 2-6, then Robin has two different ways to match up (2-6 OR 6-2). The probability of that occurring would be (2/6)(1/6) = 2/36 = 1/18

Accounting for all possible outcomes, the probability of Pam and Robin rolling the same result would be:
(1/6)(1/36) + (5/6)(1/18) =
1/216 + 5/108 =
1/216 + 10/216 =
11/216

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Mon Dec 17, 2018 12:52 pm

alanforde800Maximus wrote:Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?

A. 1/216

B. 1/36

C. 5/108

D. 11/216

E. 1/18

Very nice problem!
$$? = {{\# \,\,{\rm{favorables}}} \over {\# \,\,{\rm{total}}\,\,\left( {{\rm{equiprobables}}} \right)}}$$

$$\# \,\,{\rm{total}}\,\,\left( {{\rm{equiprobables}}} \right) = {6^2} \cdot {6^2}\,\,\,\,\left( {{\rm{taking}}\,\,{\rm{results}}\,\,{\rm{in}}\,\,{\rm{order,}}\,\,{\rm{for}}\,\,{\rm{both}}\,\,{\rm{players}}} \right)$$
$$\# \,\,{\rm{favorables}}\,\,\, = \,\,\,\left\{ \matrix{
\,6\,\,::\,\,{\rm{first}}\,\,{\rm{player}}\,\,{\rm{gets}}\,\,{\rm{same}}\,\,{\rm{number}}\,\,{\rm{twice,}}\,\,{\rm{second}}\,\,{\rm{player}}\,\,{\rm{gets}}\,\,{\rm{same}}\,\,{\rm{ones}} \hfill \cr
\,\,\, + \,\,\,\,\,\left( {{\rm{mutually}}\,\,{\rm{exclusive}}\,{\rm{!}}} \right) \hfill \cr
\,\left( {36 - 6} \right) \cdot 2!\,\,\,::\,\,\,{\rm{first}}\,\,{\rm{gets}}\,\,\left( {x,y} \right)\,\,{\rm{with}}\,\,x \ne y\,\,,\,\,{\rm{second}}\,\,{\rm{gets}}\,\,\left( {x,y} \right)\,\,{\rm{or}}\,\,\left( {y,x} \right) \hfill \cr} \right.$$

$$? = {{66} \over {{6^4}}} = {{11} \over {{6^3}}} = {{11} \over {216}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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swerve: Legendary Member; Posts: 2545; Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Tue Dec 18, 2018 8:16 am

There are 2 cases:

1. Both dices show the same number, so it's 6 out of 36. Then for the other player, the probability to roll the same number is (1/6)*(1/6).
2. There are 30 out of 36 options for the dices to be different numbers. The probability for the other player to roll the same is (2/6)*(1/6).

Probability = (1/6)*(1/6)*(1/6) + (5/6)*(2/6)*(1/6) = 11/216.

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by Scott@TargetTestPrep » Sun Mar 03, 2019 6:35 pm

alanforde800Maximus wrote:Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?

A. 1/216

B. 1/36

C. 5/108

D. 11/216

E. 1/18

Pam can obtain either of: (1) a pair of the same two numbers, or (2) a pair of two different numbers. We have to consider each case separately because the probability of Robin matching is different for each case. Here's why: If Pam rolls a (2,2), for example, then Robin has only 1 chance in 36 of matching. But if Pam rolls a (6,2), for example, then Robin can match by rolling either a (6,2) or a (2,6), so the probability is 2/36, or 1/18.

If Pam rolls a pair of same two numbers, she has a probability of 1 x 1/6 = 1/6 of rolling them. Then Robin has a probability of 1/6 x 1/6 = 1/36 to match them. So the probability they will roll the same set of two numbers when Pam rolls a pair of same numbers is 1/6 x 1/36 = 1/216.

If Pam rolls a pair of different numbers, she has a probability of 1 x 5/6 = 5/6 of rolling them. Then Robin has a 1/6 x 1/6 = 1/36 chance for her first number to match Pam's first number and her second number to match Pam's second number, and another 1/6 x 1/6 = 1/36 chance for her first number to match Pam's second number and her second number to match Pam's first number. Therefore, Robin has a 2/36 chance to match Pam's numbers. So the probability they will roll the same set of two numbers when Pam rolls a pair of different numbers is 5/6 x 2/36 = 10/216.

Therefore, the probability that they will roll the same set of two numbers 1/216 + 10/216 = 11/216.

Alternate Solution:

Suppose Pam rolls the first die. There are two cases: a) Robin's first die will match Pam's first die (a probability of 1/6) or b) Robin's first die will not match Pam's first die (a probability of 5/6).

In the first case, the only way they will get the same set of numbers is if Robin's second roll matches Pam's second roll; which has a probability of 1/6. Since the probability of this case was 1/6, the probability of obtaining a matched pair from this scenario is 1/6 x 1/6 = 1/36.

In the second case, to obtain a matched pair, Pam's second roll must match Robin's first roll (a probability of 1/6) and Robin's second roll must match Pam's first roll (a probability of 1/6). Since the probability of this case was 5/6, the probability of obtaining a matched pair from this scenario is 1/6 x 1/6 x 5/6 = 5/216.

We note that the two scenarios are mutually exclusive events. Therefore, the total probability of obtaining a matched pair is simply the sum of the individual probabilities of obtaining a matched pair from each case, which is 1/36 + 5/216 = 6/216 + 5/216 = 11/216.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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