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Brian plays a game in which two fair, six-sided dice are rol

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Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

A. 11/12
B. 7/12
C. 1/2
D. 5/12
E. 1/3
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Jun 2014, 08:55, edited 1 time in total.
Renamed the topic and edited the question.

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sidinsin wrote:
Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

A. 11/12
B. 7/12
C. 1/2
D. 5/12
E. 1/3


Brian will win in cases when both dice show an even number as well as in cases when an odd number on one die is less than an even number on another.

A. An even number on both dice can occur in 3*3 = 9 cases.

B. An odd number on one die is less than an even number on another:

(1, 2) and (2, 1)

(1, 4) and (4, 1)

(1, 6) and (6, 1)

(3, 4) and (4, 3)

(3, 6) and (6, 3)

(5, 6) and (6, 5)

12 cases here.

P = favorable/total = (9 + 12)/36 = 7/12.

Answer: C.
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Re: Brian plays a game in which two fair, six-sided dice are rol [#permalink]

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New post 18 Jun 2014, 09:59
What does 'die' means here?

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Re: Brian plays a game in which two fair, six-sided dice are rol [#permalink]

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New post 18 Jun 2014, 10:14
Hi Bunuel
Could you clarify, "Brian will win in cases when both dice show an even number as well as in cases when an odd number on one die is less than an even number on another."

The question stem does not say anything about the odd number has to be less than the even number. IMO, the winning combinations are:

1.) e-e: 3/6*3/6
2) o-e
1,2:1,4:1:6
3,2: 3,4: 3,6
5,2: 5,4: 5,6

Winning prb: 3/6*3/6+ (9/36)*2= 1/4+ 1/2= 3/4

Thanks

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sidinsin wrote:
Hi Bunuel
Could you clarify, "Brian will win in cases when both dice show an even number as well as in cases when an odd number on one die is less than an even number on another."

The question stem does not say anything about the odd number has to be less than the even number. IMO, the winning combinations are:

1.) e-e: 3/6*3/6
2) o-e
1,2:1,4:1:6
3,2: 3,4: 3,6
5,2: 5,4: 5,6

Winning prb: 3/6*3/6+ (9/36)*2= 1/4+ 1/2= 3/4

Thanks


For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money.

So, for example:
In case of (2=even, 4=even) he wins 2 + 4 = $6.
In case of (2=even, 5=odd) he losses 2 - 5 = $3.
In case of (6=even, 1=odd) he wins 6 - 1 = $5.

Hope it's clear.
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New post 18 Jun 2014, 10:24
Thanks make sense. Need to read the stem more carefully.

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Brian plays a game in which two fair, six-sided dice are rol [#permalink]

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Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

A) 11/12
B) 7/12
C) 1/2
D) 5/12
E) 1/3

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Re: Brian plays a game in which two fair, six-sided dice are rol [#permalink]

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New post 23 Jul 2014, 22:14
maggie27 wrote:
Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

A) 11/12
B) 7/12
C) 1/2
D) 5/12
E) 1/3



Please read the rules before posting. This question is already discussed over here
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Brian plays a game in which two fair, six-sided dice are rol [#permalink]

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New post 28 Jul 2014, 03:54
Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

11/12
7/12
1/2
5/12
1/3

Can someone please explain

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New post 28 Jul 2014, 04:08
Vijayeta wrote:
Brian plays a game in which two fair, six-sided dice are rolled simultaneously. For each die, an even number means that he wins that amount of money and an odd number means that he loses that amount of money. What is the probability that Brian makes money on his first roll?

11/12
7/12
1/2
5/12
1/3

Can someone please explain


Merging topics. Please refer to the discussion above.

P.S. Please search before posting and put A, B, C, D, and E in front of the answer choices.
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Re: Brian plays a game in which two fair, six-sided dice are rol [#permalink]

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This is an interesting question and needs you to read the question stem attentively.
We are sure that Brian will win in the case when there is an even number and loose the same amount if there is an odd number.

Bifurcating the cases.

1. Brian will win some money if there are even numbers on both the dices.
2. Brian will win some money if the even number on one die is more than the odd number. Here Brian will win the money equal to the difference of the two numbers on the dices.

Case 1:
Even numbers on both dices.
Since there are 3 even numbers on a dice: 2, 4, 6
This is possible in 3*3 = 9 ways

Case 2:
Even number is greater the odd number
This is possible in the following ways
(2, 1), (4, 1),(4, 3), (6, 1), (6, 3), (6, 5)
These numbers can be interchanged on both the dices.
Hence a total of 12 cases.

Total outcomes = 6*6 = 36
Probability = 12+9/36 = 21/36 = 7/12

Option B.
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Re: Brian plays a game in which two fair, six-sided dice are rol   [#permalink] 25 Sep 2017, 12:09
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