It is currently 24 Nov 2017, 00:51

Close

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
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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Daniel is playing a dice game in which a dice is rolled 5 times

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

9 KUDOS received
Director
Director
User avatar
Joined: 10 Mar 2013
Posts: 591

Kudos [?]: 482 [9], given: 200

Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
GMAT ToolKit User
Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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

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

Source: PerfectScores
[Reveal] Spoiler: OA

_________________

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 !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660


Last edited by BrainLab on 21 Sep 2015, 04:13, edited 1 time in total.

Kudos [?]: 482 [9], given: 200

7 KUDOS received
Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2672

Kudos [?]: 1776 [7], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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

C is the correct answer.

Kudos [?]: 1776 [7], given: 794

Director
Director
User avatar
G
Joined: 23 Jan 2013
Posts: 603

Kudos [?]: 27 [0], given: 41

Schools: Cambridge'16
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 21 Sep 2015, 01:38
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?

Kudos [?]: 27 [0], given: 41

Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2672

Kudos [?]: 1776 [0], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 21 Sep 2015, 03:01
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.

Kudos [?]: 1776 [0], given: 794

Senior Manager
Senior Manager
avatar
Joined: 27 Dec 2013
Posts: 303

Kudos [?]: 39 [0], given: 113

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

Show Tags

New post 21 Sep 2015, 03:36
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.

_________________

Kudos to you, for helping me with some KUDOS.

Kudos [?]: 39 [0], given: 113

Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2672

Kudos [?]: 1776 [0], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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.

Kudos [?]: 1776 [0], given: 794

Manager
Manager
avatar
Joined: 29 Jul 2015
Posts: 159

Kudos [?]: 191 [0], given: 59

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

Show Tags

New post 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 ?

Kudos [?]: 191 [0], given: 59

Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2672

Kudos [?]: 1776 [0], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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.

Kudos [?]: 1776 [0], given: 794

Current Student
avatar
Joined: 09 Aug 2015
Posts: 94

Kudos [?]: 30 [0], given: 7

GMAT 1: 770 Q51 V44
GPA: 2.3
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 21 Sep 2015, 05:29
Good question. Forgot to add the case where the next 3 could be identical and not 4s

Kudos [?]: 30 [0], given: 7

Director
Director
User avatar
Joined: 10 Mar 2013
Posts: 591

Kudos [?]: 482 [0], given: 200

Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
GMAT ToolKit User
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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 !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Kudos [?]: 482 [0], given: 200

2 KUDOS received
Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2672

Kudos [?]: 1776 [2], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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)

Hope this helps.

Kudos [?]: 1776 [2], given: 794

Director
Director
User avatar
Joined: 10 Mar 2013
Posts: 591

Kudos [?]: 482 [0], given: 200

Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
GMAT ToolKit User
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 21 Sep 2015, 08:08
Thanks engr2012, yes this information helped me. I realised my mistake in the moment I've written my question.
_________________

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 !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Kudos [?]: 482 [0], given: 200

1 KUDOS received
Manager
Manager
avatar
S
Joined: 13 Mar 2013
Posts: 179

Kudos [?]: 75 [1], given: 25

Location: United States
Concentration: Leadership, Technology
GPA: 3.5
WE: Engineering (Telecommunications)
Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 21 Sep 2015, 11:20
1
This post received
KUDOS
Hi Bunuel ,

please solve this question . Please did not understand the above solution .
_________________

Regards ,

Kudos [?]: 75 [1], given: 25

6 KUDOS received
Intern
Intern
avatar
Joined: 14 Jul 2015
Posts: 6

Kudos [?]: 29 [6], given: 9

Location: United States
Concentration: Strategy, Operations
GPA: 3.6
WE: Engineering (Energy and Utilities)
GMAT ToolKit User
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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 :)

Kudos [?]: 29 [6], given: 9

1 KUDOS received
Manager
Manager
avatar
Joined: 29 Jul 2015
Posts: 159

Kudos [?]: 191 [1], given: 59

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

Show Tags

New post 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.

Kudos [?]: 191 [1], given: 59

4 KUDOS received
Intern
Intern
avatar
Joined: 14 Jul 2015
Posts: 6

Kudos [?]: 29 [4], given: 9

Location: United States
Concentration: Strategy, Operations
GPA: 3.6
WE: Engineering (Energy and Utilities)
GMAT ToolKit User
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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 :)

Kudos [?]: 29 [4], given: 9

Intern
Intern
avatar
Joined: 10 Mar 2014
Posts: 24

Kudos [?]: 3 [0], given: 299

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

Show Tags

New post 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

Kudos [?]: 3 [0], given: 299

Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2672

Kudos [?]: 1776 [0], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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.

Kudos [?]: 1776 [0], given: 794

Manager
Manager
avatar
Joined: 02 Jul 2015
Posts: 109

Kudos [?]: 35 [0], given: 58

Schools: ISB '18
GMAT 1: 680 Q49 V33
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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)

Sum equals: 96/216

Whats wrong with this approach?

Kudos [?]: 35 [0], given: 58

Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2672

Kudos [?]: 1776 [0], given: 794

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

Show Tags

New post 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.

Kudos [?]: 1776 [0], given: 794

Re: Daniel is playing a dice game in which a dice is rolled 5 times   [#permalink] 02 Oct 2015, 04:30

Go to page    1   2    Next  [ 32 posts ] 

Display posts from previous: Sort by

Daniel is playing a dice game in which a dice is rolled 5 times

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.