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# Three dice (each having six faces with each face having one number

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Joined: 27 May 2014
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GMAT 1: 730 Q49 V41
Three dice (each having six faces with each face having one number  [#permalink]

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05 Feb 2017, 23:06
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Difficulty:

45% (medium)

Question Stats:

63% (01:29) correct 37% (01:24) wrong based on 103 sessions

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Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?
(A) 36
(B) 81
(C) 91
(D) 112
(E) 116

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Re: Three dice (each having six faces with each face having one number  [#permalink]

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06 Feb 2017, 00:56
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saswata4s wrote:
Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?
(A) 36
(B) 81
(C) 91
(D) 112
(E) 116

Hi

#1 Direct counting:

one 2 + two 2's + three 2's = 1*5*5*3!/2! + 1*1*5*3!/2! + 1*1*1 = 75 + 15 + 1 = 91

# Indirect counting:

Total # of outcomes - # of outcomes without 2's = $$6^3 - 5^3 = 91$$

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Re: Three dice (each having six faces with each face having one number  [#permalink]

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06 Feb 2017, 07:54
2
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saswata4s wrote:
Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?
(A) 36
(B) 81
(C) 91
(D) 112
(E) 116

Number of possible outcomes such that at least one dice shows the number 2 = Total Outcomes - All outcomes without any 2

Total Outcomes on three dice = 6*6*6 = 216

Total Outcomes Without any 2 = 5*5*5 (as all 5 outcomes except "2" on each die is acceptable) = 125

Required Outcomes = 216 - 125 = 91

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Re: Three dice (each having six faces with each face having one number  [#permalink]

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08 Feb 2017, 18:48
saswata4s wrote:
Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?
(A) 36
(B) 81
(C) 91
(D) 112
(E) 116

We need to determine, when rolling three dice, the number of outcomes in which at least one die shows a 2.

We should first recall the following formula:

Total number of ways to roll 3 dice = number of ways with at least one 2 + number of ways with no 2s.

Since there are 6 x 6 x 6 = 216 possible outcomes when rolling 3 dice, we have:

216 = number of ways with at least one 2 + number of ways with no 2s

number of ways with at least one 2 = 216 - number of ways with no 2s

There are 5 x 5 x 5 = 125 outcomes in which no 2 is shown.

Thus, there are 216 - 125 = 91 outcomes in which at least one 2 is shown.

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Re: Three dice (each having six faces with each face having one number  [#permalink]

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05 May 2017, 08:14
For 1 time 2 -

1*5*5*3 = 75

For 2 times = 1*1*5*3 = 15

For 3 times = 1 = 75+15+1 = 91.
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Joined: 20 Oct 2018
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Three dice (each having six faces with each face having one number  [#permalink]

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30 Dec 2018, 10:58
vitaliyGMAT wrote:
saswata4s wrote:
Three dice (each having six faces with each face having one number from 1 to 6) are rolled. What is the number of possible outcomes such that at least one dice shows the number 2?
(A) 36
(B) 81
(C) 91
(D) 112
(E) 116

Hi

#1 Direct counting:

one 2 + two 2's + three 2's = 1*5*5*3!/2! + 1*1*5*3!/2! + 1*1*1 = 75 + 15 + 1 = 91

# Indirect counting:

Total # of outcomes - # of outcomes without 2's = $$6^3 - 5^3 = 91$$

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Why do we divide by 2! In the direct counting method?
Three dice (each having six faces with each face having one number &nbs [#permalink] 30 Dec 2018, 10:58
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