Wu rolls two fair, six-sided dice. What is the probability that Wu

Question

Wu rolls two fair, six-sided dice. What is the probability that Wu rolls at least one five but no sixes?

A. 5/36
 
B. 275/1296
 
C. 2/9
 
D. 1/4
 
E. 5/18

joepc · Accepted Answer

Answer is D :- 1/4

Approach :-

Cases to get at least one 5 and No sixes

Case 1 :-  5_(this blank can be filled by 1,2,3,4 except 5 and 6) hence 4 ways

Case 2 :-  _5(this blank can be filled by 1,2,3,4 except 5 and 6) hence 4 ways

Case 3:-   5  5  = 1 ways

Favorable Outcomes :-  Case 1 + Case 2 + Case 3 = 9 ways
 
Total Outcomes  = 36

Probability(Wu rolls at least one five but no sixes) = 9/36 = 1/4

ScottTargetTestPrep · Answer

We can can break this up into scenarios:

Scenario 1:

P(5 on the first die and 1,2,3, or 4 on the second die)

1/6 x 4/6 = 4/36 = 1/9

Scenario 2:

P(1,2,3, or 4 on the first die and 5 on the second die)

4/6 x 1/6 = 4/36 = 1/9

Scenario 3:

P(5 on both dies)

1/6 x 1/6 = 1/36

Thus, the final probability is:

1/9 + 1/9 + 1/36 = 4/36 + 4/36 + 1/36 = 9/36 = 1/4.

Answer: D

sravankumar0911 · Answer

Total sample size = 36, if on the first dice only 5 comes after rolling then in the second dice we will have 5 possibilities 1,2,3,4,5(no sixes as per the question) same will be repeated with second dice. Total possibilities 10, hence probability is 10/36 or 5/18 correct answer E

Sent from my iPhone using GMAT Club Forum

Sent from my iPhone using GMAT Club Forum

mohshu · Answer

u r counting the possibility of 5,5 two times....

favourable outcomes will 9

Vinayak Shenoy · Answer

You have to subtract 1 from 10 possibilities cos you are counting 5,5 twice. Actual no of possibilities is 9 out of 36 which is 1/4 D

sravankumar0911 · Answer

Got it thanks for pointing out the mistake

Sent from my iPhone using GMAT Club Forum

BillyZ · Answer

Official solution from Veritas Prep.

As is often the case with probability problems, there are multiple approaches to finding the right answer.

One such method is to directly count both the good outcomes and the total outcomes for rolling two fair, six-sided dice.

The total outcomes are straightforward: each die has six possible outcomes, so there are  6∗6=36  total outcomes.

The good outcomes are those involving either:

a 5 on the first roll and a non-5, non-6 on the second roll – 4 such outcomes, from (5,1) through (5,4); or
a non-5, non-6 on the first roll and a 5 on the second roll – 4 such outcomes, from (1,5) through (4,5); or 
a 5 on the first roll and a 5 on the second roll – one such outcome (5,5).

This is an “or” case, so the total good outcomes are  4+4+1=9 .Overall, the probability is  \frac{9}{36}=\frac{1}{4} , and the correct answer is D.

KrishnakumarKA1 · Answer

Way to get atleast one 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6). 
Total possible cases = 11. Since we dont want any 6, we remove (5,6), (6,5). Hence the total cases will be 9.
number of possibilities = 6*6 = 36

Therefore probability = 9/36 = 1/4. 
Option D

nks2611 · Answer

probability of getting at least one 5 and no six will be such as :

1/6*5/6 + 1/6*5/6 - 1/6*1/6= 9/36 =1/4

can someone clarify my method

Getsome2 · Answer

Thank you, all explanations above are helpful and make sense. FWIW, another way to think about it is noticing in the question "at least" - which to me rings a bell to try using the complement to solve.

So looking at the first element of "at least one 5" the complement would be the probability of not rolling a 5 on either roll:

5/6 (1,2,3,4 and 6 out of 6 possibilities) times 5/6 (same logic as first roll) which gets you 25/36; this represents the probability of NOT rolling a 5 on either/both rolls

So then subtracting 25/36 from 1 gets you 11/36 or the probability of  just  rolling a 5 on at least one dice

Now the 2nd element of rolling no 6's - again thinking about the complement of how you can roll a 6: there are just three ways you can roll a 6 using two dice: 1) one 6 on the first and no 6 on the second 2) no six on the first and one 6 on the second or 3) 6's on both dice

However, we know from the first part of the problem that we're looking for the probability of rolling at least one 5, so the last possibility of the 2nd element (rolling two 6's) can't be an outcome because then the first element wouldn't be satisfied

Therefore, we only have two ways for which we can roll a 6.

So to tie it all together we have 11 ways of rolling a 5 on at least one dice, however these 11 ways include the possibility of rolling a 6 on one of the dice that is NOT a 5. Thus we must subtract out these 2 ways which gets us 9 possible ways of rolling a 5 on at least one dice (element 1) and no 6's (element 2)

9/36 simplifies to 1/4

D is your answer

Mugdho · Answer

There would be no 6. So we for at least we can write use reverse method- we will not consider 5 also.

Probability of having no 6 and 5 is
4/6*4/6.

So Probability of having at least 5 but no 6 will be 1-4/9=5/9

Why this method is not working?  ScottTargetTestPrep

Posted from my mobile device

ScottTargetTestPrep · Answer

The probability of no 5 and no 6 is indeed 4/6 * 4/6, however, the complement of "no 5 and no 6" is not "at least one 5 and no 6". The complement of "no 5 and no 6" is "at least one 5 or at least one 6". For instance, a roll of two 6 is not included in "no 5 and no 6", which means it is in the complement.

The complement of "at least one five and no sixes" is "at least one six or no fives". To calculate this probability, you need to calculate P(at least one six), P(no fives), and P(at least one six and no fives). Then you need to apply:

P(at least one six or no fives) = P(at least one six) + P(no fives) - P(at least one six and no fives).

After you calculate the above probability, you need to subtract it from 1. I think that's a lot of work given that you can calculate the same probability much faster if you use the direct approach.

BrushMyQuant · Answer

Given that Wu rolls two fair, six-sided dice and We need to find What is the probability that Wu rolls at least one five but no sixes?

As we are rolling two dice => Number of cases =  6^2  = 36

Let's list down the cases in which in the first roll we got a 5 and in second roll we got numbers apart from 6
(5,1), (5,2), (5,3), (5,4), (5,5) => 5 cases

Let's list down the cases in which in the second roll we got a 5 and in first roll we got numbers apart from 6
(1,5), (2,5), (3,5), (4,5), (5,5), note that (5,5) is already considered above => 4 cases here

=> Total cases = 5 + 4 = 9

=>  P(Wu rolls at least one five but no sixes)  =  \frac{9}{36}  =  \frac{1}{4}

So,  Answer will be D 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

PSKhore · Answer

Total ways = 6*6 = 36

Probability of rolling at least one five but no sixes:
(1,5),(2,5),(3,5),(4,5),(5,5)

Similar possibilities for the other dice, except (5,5) because it will be counted twice.

5*2 - 1
= 9

9/36
=1/4

Wu rolls two fair, six-sided dice. What is the probability that Wu

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Wu rolls two fair, six-sided dice. What is the probability that Wu

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards