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:

Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

20 Sep 2015, 14:18

9

This post received KUDOS

29

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

27% (02:20) correct
73% (02:16) wrong based on 466 sessions

HideShow timer Statistics

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ? Note: the dice is an unbiased die with numbers 1 to 6

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

20 Sep 2015, 16:01

7

This post received KUDOS

14

This post was BOOKMARKED

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 04:37

shriramvelamuri wrote:

Hi Engr2012,

Please help me with this question. I did the following way..

4 4 X X X (X is the probability of obtaining another 4)

Attempt 1:

1st X = 4. the probability is 1/6 (Game Won- End of Game)

Attempt 2:

2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won)

Attempt 3

3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216

Total probability for 4= 91/ 216.

The three XXX are a number (Say 5) probability = 1/216.

I am getting 92/216.

Where am I going wrong here.

Please explain kindly.

Engr2012 wrote:

Temurkhon wrote:

to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2

to have any number except 4 is 5/6*1/6*1/6=5/216

5/216+1/2=5/216+108/216=113/216

what is wrong?

You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done.

Look at my solution above.

Hope this helps.

Your assumption that the game "stops" once you get 3 of 4s is not correct. You still need to count for all 5 cases and then evaluate whether the person has won. Look at my solution above for scenario 1.

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 04:55

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ? Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216

Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216

I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ?

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 04:58

kunal555 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ? Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216

Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216

I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ?

Nowhere is it mentioned that the game is stopped as soon he has won the game. Thus you still need to consider the other cases and not assume things that are not given.

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 07:51

Engr2012 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.

Hi Engr2012,

I've also set up 2 correct scenarios, but had problenbs calculating them - particularly the second scenario: 1] (1/6)*(5/6)*(5*6)*3!/2!=75/216 --> Ok, here no questions Not 4*Not 4*4 (1/6)

BUT

2] 4 4 N N N --> 5/6*5/6*5/6 Not4*Not4*Not4, why do we have to choose 1/6 twice after 5/6 ?
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 08:02

2

This post received KUDOS

BrainLab wrote:

Engr2012 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.

Hi Engr2012,

I've also set up 2 correct scenarios, but had problenbs calculating them - particularly the second scenario: 1] (1/6)*(5/6)*(5*6)*3!/2!=75/216 --> Ok, here no questions Not 4*Not 4*4 (1/6)

BUT

2] 4 4 N N N --> 5/6*5/6*5/6 Not4*Not4*Not4, why do we have to choose 1/6 twice after 5/6 ?

That is because once you have 1 particular number out of (1,2,3,5,6), in order to "win" you MUST get that 1 particular number twice more in order to "win". Selecting 1 number out of 6 possible = 1/6. You can select 1 particular number of out of the remaining 5 numbers iwth probability of 5/6.

The winning combinations can be 44555 or 44111 or 44222 or 44333 or 44666

Thus for 4 4 N N N , in which N \(\neq\)4 = (5/6)*(1/6)*(1/6)

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 13:03

6

This post received KUDOS

kunal555 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ? Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216

Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216

I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ?

Hi Kunal555,

I was also having the same question as you have. But look at the details of question. It says exactly 3 times. If it is not exactly 3 times, he will lose the game. Hence, the game should go on for total 5 throws, then only we can decide whether he won the game or not. If we stop the game at the third throw when he gets 4, then we are eliminating his chance of losing the game because we don't know the result of 4th and 5th throw. If he gets 4 in the fourth throw also, then it means he will lose the game because now, the total number of 4s are not exactly 3.

Please award kudos if you agree with my explanation.
_________________

Please award kudos if you like my explanation. Thanks

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 13:37

1

This post received KUDOS

abhigoyal1989 wrote:

kunal555 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ? Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Probability of getting 4 in first throw = 1/6 probability of getting 4 in second throw = 5/6*1/6 = 5/36 Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216 Total probability of getting 1 more 4 will be 1/6+5/36+25/216 = 91/216

Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216 there are 5 such numbers left. so probability will be 5*1/216 = 5/216 Total probability will be 91/216+5/216 = 97/216

I dont understand why does he have to throw the dice in case of 4, after the game is already won. The question does not specify whether he throws the dice after winning the game. Logically he shouldn't. So, to consider or not to consider the events after winning the game in such questions ?

Hi Kunal555,

I was also having the same question as you have. But look at the details of question. It says exactly 3 times. If it is not exactly 3 times, he will lose the game. Hence, the game should go on for total 5 throws, then only we can decide whether he won the game or not. If we stop the game at the third throw when he gets 4, then we are eliminating his chance of losing the game because we don't know the result of 4th and 5th throw. If he gets 4 in the fourth throw also, then it means he will lose the game because now, the total number of 4s are not exactly 3.

Please award kudos if you agree with my explanation.

You are right. Also, first statement says "Daniel is playing a dice game in which a dice is rolled 5 times." IMO it is simple and clear that he plays a game in which a dice HAS TO BE ROLLED 5 TIMES. And if in 5 trials, a number shows up more than 3 times, he will loose.

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

21 Sep 2015, 16:13

4

This post received KUDOS

shriramvelamuri wrote:

Hi Engr2012,

Please help me with this question. I did the following way..

4 4 X X X (X is the probability of obtaining another 4)

Attempt 1:

1st X = 4. the probability is 1/6 (Game Won- End of Game)

Attempt 2:

2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won)

Attempt 3

3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216

Total probability for 4= 91/ 216.

The three XXX are a number (Say 5) probability = 1/216.

I am getting 92/216.

Where am I going wrong here.

Please explain kindly.

Engr2012 wrote:

Temurkhon wrote:

to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2

to have any number except 4 is 5/6*1/6*1/6=5/216

5/216+1/2=5/216+108/216=113/216

what is wrong?

You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done.

Look at my solution above.

Hope this helps.

Hi Shriram,

You are missing one crucial information that he has to throw dice 5 times. and he has to get same number exactly 3 times. If you are stopping at "attempt 1" where you are saying game will end and he will won because he has thrown same number 3 times, you are missing "exactly" term in the question. Thus to ensure he won the game, he has to throw a number other than 4. That is why we multiply 5/6 two times (4th throw and 5th throw). Your logic would have been fine in case the question says "he will win as soon as he gets three times a same number"

Please award kudos if you like my explanation. Thanks
_________________

Please award kudos if you like my explanation. Thanks

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

25 Sep 2015, 08:07

Engr2012 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.

Excellent explanation! But just out of curiosity, can this problem also be solved using combinatorics approach? Thanks

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

25 Sep 2015, 12:18

Rk91 wrote:

Engr2012 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216 b) 75/216 c) 80/216 d) 90/216 e) 100/216

Source: PerfectScores

Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.

Excellent explanation! But just out of curiosity, can this problem also be solved using combinatorics approach? Thanks

The solution above does use combinatorics with counting principle used to come up with the final probability. I dont think pure combinatorics method will be time effective strategy for such questions.

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

02 Oct 2015, 04:23

Bunuel/ Karishma/ all:

Can you please provide an explanation for the question here, as my answer is not even present as a choice!

Explanation: game gets over after getting a unique number 3 times:

1. 444: 1/6 (can be written as 1/6*6/6*6/6 for calculations= 36/216) 2. 44N4: 5/6*1/6 (can be written as 5/6*1/6*6/6= 30/216) 3. 44NN4: 5/6*5/6*1/6 (25/216) 4. 44NNN:5/6*1/6*1/6 (5/216)

Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

02 Oct 2015, 04:30

longfellow wrote:

Bunuel/ Karishma/ all:

Can you please provide an explanation for the question here, as my answer is not even present as a choice!

Explanation: game gets over after getting a unique number 3 times:

1. 444: 1/6 (can be written as 1/6*6/6*6/6 for calculations= 36/216) 2. 44N4: 5/6*1/6 (can be written as 5/6*1/6*6/6= 30/216) 3. 44NN4: 5/6*5/6*1/6 (25/216) 4. 44NNN:5/6*1/6*1/6 (5/216)

Sum equals: 96/216

Whats wrong with this approach?

The mistake you are doing is to assume that the game stops as soon as you get 3 is a row or even when you get a number 3 times out of 4. There will be 5 throws of the dice no matter what. Look at my solution above.