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# Ps - Probablity

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Senior Manager
Joined: 15 Apr 2005
Posts: 414

Kudos [?]: 17 [0], given: 0

Location: India, Chennai

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05 Jan 2006, 04:21
00:00

Difficulty:

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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Find the probability of getting a larger number than the previous number every time, when an unbiased die is thrown three times.

(a) 1/18 (b) 15/216

(c) 25/216 (d) 5/54

Kudos [?]: 17 [0], given: 0

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5036

Kudos [?]: 436 [0], given: 0

Location: Singapore

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05 Jan 2006, 04:47
Possibilites:

First throw is 1: 2,3 - 2,6; 3,4 - 3,6; 4,5 - 4,6; 5,6 --> Total: 10 ways

First throw is 2: 3,4 - 3,6; 4,5 - 4,6, 5,6 --> Total: 6 ways

First throw is 3: 4,5 - 4,6, 5,6 --> Total: 3 ways

First throw is 4: 5,6 -> Total: 1 ways

First throw is 5: No way

First throw is 6: No way

Total number of possibilites = 20

Total number of combinations with three rolls of a die = 6*6*6 = 216

Probability = 20/216 = 10/108 = 5/54 (D)

P.S: This method may not look elegant, but it's extremely fast.

Kudos [?]: 436 [0], given: 0

Director
Joined: 04 Oct 2005
Posts: 582

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

Location: Chicago

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07 Jan 2006, 00:55
5/54..

Here's my working

Prob of getting a number on the first throw= 1/6

Prob of getting a no higher on the second throw = 5C1/6 ( Only 5 nos r left)=5/6

Prob of getting a no higher than 3rd throw= 4C1/6 (Only 4 r left_=4/6

1/6*5/6*4/6=5/54

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

07 Jan 2006, 00:55
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