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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

Dice

by shashank.ism » Sat Feb 06, 2010 1:48 pm

A dice is rolled three times. Find the probability of getting a larger number than previous number each time.

a) 5/54
b) 7/54
c) 11/54
d) 13/54
e) 17/54

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Source: Beat The GMAT — Problem Solving | Sat Feb 06, 2010 1:48 pm

harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

by harsh.champ » Sat Feb 06, 2010 2:03 pm

shashank.ism wrote:A dice is rolled three times. Find the probability of getting a larger number than previous number each time.

a) 5/54
b) 7/54
c) 11/54
d) 13/54
e) 17/54

Total no. of outcomes = 6x6x6.
Total no. of favorable outcomes can be calculated as follows:-
The first number can be 1,2,3,4.
Corresponding to 1,rest 2 numbers can be selected in 5C2 ways. [ 2 out of 2,3,4,5,6]
Corresponding to 2,rest 2 numbers can be selected in 4C2 ways. [ 2 out of 3,4,5,6]
Corresponding to 3,rest 2 numbers can be selected in 3C2 ways. [ 2 out of 4,5,6]
Corresponding to 4,rest 2 numbers can be selected in 2C2 (which is 1)ways. [ 2 out of 5,6]

Hence,total no. of favorable outcomes = 5C2 + 4C2 + 3C2 + 2C2
= 10 + 6 + 3 + 1
= 20

Hence,the [spoiler]ans. is 20/216 = 5/54.[/spoiler]

IMO A.

Last edited by harsh.champ on Sat Feb 06, 2010 2:51 pm, edited 1 time in total.

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DeepthiRajan: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Sat Feb 06, 2010 1:55 pm; Thanked: 5 times

by DeepthiRajan » Sat Feb 06, 2010 2:12 pm

The way I have approached the problem is:

The sample space is {216} [6*6*6]

And the event of each new number being greater than the previous number on the dice is:

(123), (124),(125)(126)
(134),(135),(136)
(145),(146)
(156)

(234),(235),(236),(245),(246),(256)

(345),(346),(356)

(456)

There are 20 events in all which means P(E) = 20/ 216 ===> 5/54 , answer is A

If anyone can suggest a faster solution, it would be helpful.

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 am; Location: India; Thanked: 64 times; Followed by:6 members; GMAT Score:760

by harsh.champ » Sat Feb 06, 2010 2:55 pm

DeepthiRajan wrote:The way I have approached the problem is:

The sample space is {216} [6*6*6]

And the event of each new number being greater than the previous number on the dice is:

(123), (124),(125)(126)
(134),(135),(136)
(145),(146)
(156)

(234),(235),(236),(245),(246),(256)

(345),(346),(356)

(456)

There are 20 events in all which means P(E) = 20/ 216 ===> 5/54 , answer is A

If anyone can suggest a faster solution, it would be helpful.

Hey deepthi,
I found the above method posted by me to be faster than your approach.Maybe this suits you but over here you can see that you had to write down 20 possible arrangements which probably took a lot of your time.
If suppose the no. of favorable outcomes was small,lets say 5 or 6 , then this method would have worked just fine but in case of larger no. of outcomes ,I think it is better to solve the question by the formal method. What-say??

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

by Ian Stewart » Sat Feb 06, 2010 3:05 pm

There are 6^3 outcomes in total.

Now, how many ways can we get the result we want? We just need to choose three different numbers from this list:

{1, 2, 3, 4, 5, 6}

There will be exactly one way to put our selected numbers in increasing order. So there will be 6C3 = (6*5*4)/3! = 20 ways to make an increasing sequence of numbers between 1 and 6. So the answer is 20/6^3 = 5/54.

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Eldorjon: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Tue Feb 02, 2010 8:32 am; Thanked: 14 times; Followed by:6 members; GMAT Score:760

by Eldorjon » Sat Feb 06, 2010 3:58 pm

Good method of solution!! Thanks for giving additional technique Ian!

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Sun Feb 07, 2010 2:29 am

shashank.ism wrote:A dice is rolled three times. Find the probability of getting a larger number than previous number each time.

a) 5/54
b) 7/54
c) 11/54
d) 13/54
e) 17/54

DeepthiRajan wrote:
Hey deepthi,
I found the above method posted by me to be faster than your approach.Maybe this suits you but over here you can see that you had to write down 20 possible arrangements which probably took a lot of your time.
If suppose the no. of favorable outcomes was small,lets say 5 or 6 , then this method would have worked just fine but in case of larger no. of outcomes ,I think it is better to solve the question by the formal method. What-say??

There can be a lot many ways to solve a problem in mathematics and specially if its a problem of probability.
And here I think Harsh is very right in is point of view that Deepthi's procedure was fast in this case but may get impossible if there are large number of outcomes.
Harsh you have done problem with standard approach, that is very right but from exam point of view you get very less time so you will have to adopt some other methods too.

Well for this question Ian showed us the fastest and best method, and really it gave us a new approach problem. I would have never thought of this way for problem like this, rather I could have tried harsh's or deepthi's approach.

You have given a very good solution and it will surely help to increase my score in GMAT.