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

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

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New post 06 Feb 2017, 00:06
1
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A
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C
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Difficulty:

  45% (medium)

Question Stats:

62% (01:57) correct 38% (02:00) wrong based on 111 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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New post 06 Feb 2017, 01:56
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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 = \(6^3 - 5^3 = 91\)

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

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New post 06 Feb 2017, 08:54
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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

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

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New post 08 Feb 2017, 19:48
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.

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

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New post 05 May 2017, 09:14
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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Three dice (each having six faces with each face having one number  [#permalink]

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New post 30 Dec 2018, 11:58
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 = \(6^3 - 5^3 = 91\)

Answer C.


Posted from my mobile device

Why do we divide by 2! In the direct counting method?
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Three dice (each having six faces with each face having one number   [#permalink] 30 Dec 2018, 11:58
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