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A certain dice game can only be won if, when a player throws

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Joined: 12 Oct 2012
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A certain dice game can only be won if, when a player throws [#permalink]

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New post 22 Jul 2016, 09:22
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A
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D
E

Difficulty:

  85% (hard)

Question Stats:

43% (02:03) correct 57% (03:07) wrong based on 108 sessions

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A certain dice game can only be won if, when a player throws
two fair six-sided dice, the number showing on one of the dice
is a multiple of the number showing on the other. What is the
probability that a player wins this game?

(A) 11/18
(B) 11/36
(C) 1/4
(D) 2/9
(E) 7/36

[Reveal] Spoiler:
My approach is as follows :
For first dice, with 1, there will be 6 ways,
Similarly for 2 on first dice, there will be 3 ways (2,4,6)
For 3, on first dice, there will be 2 ways (3,6)
For 4, 5 ,6 , on first dice, 1 way for each.

The answer doesn't seem to be right. Am I missing out on any possibility? :?: (
[Reveal] Spoiler: OA

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Re: A certain dice game can only be won if, when a player throws [#permalink]

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New post 22 Jul 2016, 09:55
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aditi2013 wrote:
[My approach is as follows :
For first dice, with 1, there will be 6 ways,
Similarly for 2 on first dice, there will be 3 ways (2,4,6)
For 3, on first dice, there will be 2 ways (3,6)
For 4, 5 ,6 , on first dice, 1 way for each.

The answer doesn't seem to be right. Am I missing out on any possibility? :?:


First dice with 1 --> 6 ways
First dice with 2 --> 3 ways
First dice with 3 --> 2 ways
First dice with 4, 5 and 6 --> 1 + 1 + 1 = 3 ways
Total = 14

Now the entire procedure can be repeated for second dice as well --> 14 ways

We have double counted (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6) --> 6

Required number of ways = 14 + 14 - 6 = 22 ways

Probability = 22/(6 * 6) = 22/36 = 11/18

Answer: A

Kudos [?]: 1172 [5], given: 890

Manager
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Re: A certain dice game can only be won if, when a player throws [#permalink]

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New post 22 Jul 2016, 11:21
Vyshak wrote:
aditi2013 wrote:
[My approach is as follows :
For first dice, with 1, there will be 6 ways,
Similarly for 2 on first dice, there will be 3 ways (2,4,6)
For 3, on first dice, there will be 2 ways (3,6)
For 4, 5 ,6 , on first dice, 1 way for each.

The answer doesn't seem to be right. Am I missing out on any possibility? :?:


First dice with 1 --> 6 ways
First dice with 2 --> 3 ways
First dice with 3 --> 2 ways
First dice with 4, 5 and 6 --> 1 + 1 + 1 = 3 ways
Total = 14

Now the entire procedure can be repeated for second dice as well --> 14 ways

We have double counted (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6) --> 6

Required number of ways = 14 + 14 - 6 = 22 ways

Probability = 22/(6 * 6) = 22/36 = 11/18

Answer: A


Thank you Vyshak. Its pretty clear now.

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Re: A certain dice game can only be won if, when a player throws [#permalink]

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New post 23 Jul 2017, 12:15
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Or simply just list what the question asks, i.e. pairs in which one is the factor of another:
1--pairs with (1,2,3,4,5,6)
2--pairs with (1,2,4,6)
3--pairs with (1,3,6)
4--pairs with (1,2,4)
5--pairs with (1,5)
6--pairs with (1,2,3,6)

total 22 such pairs.
p=22/36=11/18.

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Re: A certain dice game can only be won if, when a player throws [#permalink]

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New post 11 Sep 2017, 23:43
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D1-d2
1-1,2,3,4,5,6
2-1,2,4,6
3-1,3,6
4-1,2,4
5-1,5
6-1,2,3,6

DICE 1 ON THE LEFT FOLLOWED BY DASH AND THEN DICE 2 OUTCOMES FOR WIN

TOTAL = 22
CASES POSSIBLE TOTAL = 6X6=36
P = 22/36
P =11/18
OPTION A
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Re: A certain dice game can only be won if, when a player throws   [#permalink] 11 Sep 2017, 23:43
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