Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 10 Sep 2013
Posts: 6

Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
10 Sep 2013, 02:30
2
This post received KUDOS
2
This post was BOOKMARKED
Question Stats:
50% (03:18) correct
50% (01:43) wrong based on 119 sessions
HideShow timer Statistics
Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number? A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6 So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. whats wrong with what I did? OA:
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 10 Sep 2013, 02:47, edited 1 time in total.
Edited the question.



Math Expert
Joined: 02 Sep 2009
Posts: 39581

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
10 Sep 2013, 02:57
sandra123 wrote: Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number? A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6 So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. whats wrong with what I did? OA: The question asks about the probability that two of the dice show the same number but the third dice shows a different number. Total # of outcomes is 6^3; Favorable outcomes are all possible scenarios of XXY: \(C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90\), where \(C^1_6\) is # of ways to pick X (the number which shows twice), \(C^1_5\) is # of ways to pick Y (out of 5 numbers left) and \(\frac{3!}{2!}\) is # of permutation of 3 letters XXY out of which 2 X's are identical. P=Favorable/Total=90/6^3=15/36. Answer: B. Or: P(XXY)=1*1/6*5/6*3!/2! > (any, the same one, different one). Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 15 Jul 2013
Posts: 9

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
10 Sep 2013, 06:48
did I read this questions wrong? How it was written made me think that all three dice were thrown individually, and the number on the first two had to match, with the number on the third die being thrown had to be the different number. So I interpreted that the pattern had to be xxy.
Is this an easy mistake to make or is it just me?



Intern
Joined: 15 Jul 2013
Posts: 9

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
10 Sep 2013, 06:50
sandra123 wrote: Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number? A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6 So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. whats wrong with what I did? OA: To do it this way it would be 1/1 * 1/6 * 5/6 = 5/36. Then multiply by 3 to get 15/36. It is 1/1 because the first die can be any number, not a specific number.



Intern
Joined: 29 Jan 2013
Posts: 44

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
10 Sep 2013, 18:37
why are you multiplying by 3. It is clearly given that first two must be same and third must be different  i.e clearly XXY
i.e (1,1, 26), (2,2, 16 without 2) ....(6,6, 15)  total 30 favorable outcomes??
When multiplying by 3 arent you considering XYX and YXX also? Can someone explain where i went wrong?



Intern
Joined: 07 Sep 2013
Posts: 25
Concentration: Entrepreneurship
GPA: 3.52

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
11 Sep 2013, 00:22
The question talks about the "first two" rolls and the "last" roll, so I don't think that Bunuel's answer is correct. He's solving for the probability that ANY TWO are the same and the other is different.
The correct answer is 1/1 x 1/6 x 5/6 = 5/36
Where did this question come from?
t1000



Intern
Joined: 30 Nov 2012
Posts: 30

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
11 Sep 2013, 00:35
tchang wrote: The question talks about the "first two" rolls and the "last" roll, so I don't think that Bunuel's answer is correct. He's solving for the probability that ANY TWO are the same and the other is different.
The correct answer is 1/1 x 1/6 x 5/6 = 5/36
Where did this question come from?
t1000 I approached it this way too....what are we doing wrong Bunuel??



Intern
Joined: 10 Sep 2013
Posts: 6

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
11 Sep 2013, 01:28
ROFLZZZ wrote: sandra123 wrote: Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number? A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6 So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. whats wrong with what I did? OA: To do it this way it would be 1/1 * 1/6 * 5/6 = 5/36. Then multiply by 3 to get 15/36. It is 1/1 because the first die can be any number, not a specific number. ok but still can be any number out of the 6 on the die. it cant be just any number, for eg ten. Also why did you multiply 5/36 with three? first two had to be the same number, last had to be different. also, its manhattan



Senior Manager
Joined: 21 Jan 2010
Posts: 329

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
12 Sep 2013, 00:28
sandra123 wrote: Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number? A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6 So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. whats wrong with what I did? OA: Total cases = 216. Favorable cases: 112,121,211.... If you fix one two numbers as above , their can be 15 such cases. You can write it down if needed for clarity and there are 6 numbers. So fav cases= 15x 6 = 90. probability = 15/36



Intern
Joined: 07 Sep 2013
Posts: 25
Concentration: Entrepreneurship
GPA: 3.52

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
12 Sep 2013, 22:05
Bluelagoon,
The question talks about the first two rolls matching and the third doesn't
In your examples:
112 would match
but 121 would be the first and third rolls matching and 211 would be the second and third rolls matching
The prompt clearly says the first two rolls match and the third doesn't
t1000



Manager
Joined: 24 Oct 2013
Posts: 174
Location: Canada
WE: Design (Transportation)

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
20 Jun 2014, 17:00
Bunuel wrote: sandra123 wrote: Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number? A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6 So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. whats wrong with what I did? OA: The question asks about the probability that two of the dice show the same number but the third dice shows a different number. Total # of outcomes is 6^3; Favorable outcomes are all possible scenarios of XXY: \(C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90\), where \(C^1_6\) is # of ways to pick X (the number which shows twice), \(C^1_5\) is # of ways to pick Y (out of 5 numbers left) and \(\frac{3!}{2!}\) is # of permutation of 3 letters XXY out of which 2 X's are identical. P=Favorable/Total=90/6^3=15/36. Answer: B. Or: P(XXY)=1*1/6*5/6*3!/2! > (any, the same one, different one). Hope it's clear. Bunuel it specifically says FIRST two show the same number and LAST one shows different. What you did here would be an answer to 'ANY TWO showing same and the REMAINING showing different'. The questions states the order too by mentioning FIRST/LAST. Please clear me if i'm wrong.



Intern
Joined: 07 Sep 2012
Posts: 3

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
20 Jun 2014, 19:48
this is clearly a problem where the sequence matters hance the favourable outcomes are 30 for XXY excluding XyX and YXX. But a great problem to sort out basic understanding and generate debate. Kudos gauravkaushik8591 wrote: Bunuel wrote: sandra123 wrote: Three dice are thrown. What is the probability that the first two show the same number, and the last one a different number? A. 3/6 B. 15/36 C.1/216 D. 15/216 E. 1/6 So I did 1/6 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/216. then i multiply by 3 and get 15/216. whats wrong with what I did? OA: The question asks about the probability that two of the dice show the same number but the third dice shows a different number. Total # of outcomes is 6^3; Favorable outcomes are all possible scenarios of XXY: \(C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90\), where \(C^1_6\) is # of ways to pick X (the number which shows twice), \(C^1_5\) is # of ways to pick Y (out of 5 numbers left) and \(\frac{3!}{2!}\) is # of permutation of 3 letters XXY out of which 2 X's are identical. P=Favorable/Total=90/6^3=15/36. Answer: B. Or: P(XXY)=1*1/6*5/6*3!/2! > (any, the same one, different one). Hope it's clear. Bunuel it specifically says FIRST two show the same number and LAST one shows different. What you did here would be an answer to 'ANY TWO showing same and the REMAINING showing different'. The questions states the order too by mentioning FIRST/LAST. Please clear me if i'm wrong.



Manager
Joined: 24 Oct 2012
Posts: 65
WE: Information Technology (Computer Software)

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
22 Jun 2014, 01:00
Bear with me till i get my probability concepts clearer,
In this question, i had solved it this way, 1/1 * 1/6 * 5/6 (first two show the same number, the last one another number) = 5/36.
and there is no choice with this answer. any thoughts on how to go ahead in such problems?



Intern
Joined: 23 Mar 2014
Posts: 12

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
23 Jun 2014, 12:02
I approached this question the same way as many others:
First dice (can be any number) = 6/6 Second dice (must be the same as first dice) = 1/6 Third dice (any number other than the first 2 dice) = 5/6
Therefore the probability is: 6/6 * 1/6 * 5/6 = 30/216 = 5/36
I'm not understanding why this number needs to be multiplied by 3 either.



Manager
Joined: 24 Oct 2012
Posts: 65
WE: Information Technology (Computer Software)

Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
23 Jun 2014, 16:39
beef001.both of us are on same page. I donot see need to multiply by 3 either. any experts advice here please.



Math Expert
Joined: 02 Sep 2009
Posts: 39581

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
24 Jun 2014, 06:09



Manager
Joined: 13 Dec 2011
Posts: 52
GPA: 4

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
24 Jun 2014, 12:06
Another way to look at this question: total # of outcomes are 216 (6^3...6 for each dice)...now consider 6 throws in which you have sequences such as 11(2,3,4,5,6), 22(1,3,4,5,6), ..., 66(1,2,3,4,5)...so these are 6*5=30 outcomes....so your result is 30/216 which is 5/36..now multiple it by 3 for 3 possible sequences on 3 dice for the above 30 outcomes.



Intern
Joined: 14 May 2014
Posts: 19

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
09 Jul 2014, 11:41
Bunuel wrote: beef001 wrote: I approached this question the same way as many others:
First dice (can be any number) = 6/6 Second dice (must be the same as first dice) = 1/6 Third dice (any number other than the first 2 dice) = 5/6
Therefore the probability is: 6/6 * 1/6 * 5/6 = 30/216 = 5/36
I'm not understanding why this number needs to be multiplied by 3 either. This is explained in my solution above. It should be P(XXY)=1*1/6*5/6*3!/2! > (any, the same one, different one). XXY can occur in three different ways XXY, XYX, or YXX: \(\frac{3!}{2!}=3\) is # of permutation of 3 letters XXY out of which 2 X's are identical. Hi Bunuel, The confusion seems to be stemming from the question text. The question mentions that 'first two' dice should be same and the last dice should be different. So only XXY should be a success and not XYX or YXX.. So should the correct answer be 6x1x5/216 (Any number x same number x remaining numbers)/216 = 5/36? Please advise. Thanks!



Intern
Joined: 12 May 2014
Posts: 10

Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
15 Jul 2014, 21:26
The question asks about the probability that two of the dice show the same number but the third dice shows a different number.
Total # of outcomes is 6^3;
Favorable outcomes are all possible scenarios of XXY: \(C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90\), where \(C^1_6\) is # of ways to pick X (the number which shows twice), \(C^1_5\) is # of ways to pick Y (out of 5 numbers left) and \(\frac{3!}{2!}\) is # of permutation of 3 letters XXY out of which 2 X's are identical.
P=Favorable/Total=90/6^3=15/36.
Answer: B.
Or: P(XXY)=1*1/6*5/6*3!/2! > (any, the same one, different one).
Hope it's clear.[/quote]
Hello Bunuel,
Why are we calculating the permutation of the three letters which have been selected ? Shouldn't it be just 6.5.6 = 180 ? In case of permutation we are also considering the arrangement XXY, XYX, YXX. However, the question clearly states that the first two need to be identical while the last one is different ?
Any help will be appreciated. Thank you



Math Expert
Joined: 02 Sep 2009
Posts: 39581

Re: Three dice are thrown. What is the probability that the firs [#permalink]
Show Tags
16 Jul 2014, 02:37
parul1591 wrote: The question asks about the probability that two of the dice show the same number but the third dice shows a different number.
Total # of outcomes is 6^3;
Favorable outcomes are all possible scenarios of XXY: \(C^1_6*C^1_5*\frac{3!}{2!}=6*5*3=90\), where \(C^1_6\) is # of ways to pick X (the number which shows twice), \(C^1_5\) is # of ways to pick Y (out of 5 numbers left) and \(\frac{3!}{2!}\) is # of permutation of 3 letters XXY out of which 2 X's are identical.
P=Favorable/Total=90/6^3=15/36.
Answer: B.
Or: P(XXY)=1*1/6*5/6*3!/2! > (any, the same one, different one).
Hope it's clear. Hello Bunuel,
Why are we calculating the permutation of the three letters which have been selected ? Shouldn't it be just 6.5.6 = 180 ? In case of permutation we are also considering the arrangement XXY, XYX, YXX. However, the question clearly states that the first two need to be identical while the last one is different ?
Any help will be appreciated. Thank you[/quote] There are no first or second or third die there. Any die can be called first or second when thrown.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Re: Three dice are thrown. What is the probability that the firs
[#permalink]
16 Jul 2014, 02:37







