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Alejandro has a six-sided die with faces numbered 1 through
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03 Jun 2017, 02:29
03 Jun 2017, 02:29
Expert Reply
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Question Stats:
79% (00:59) correct
20% (00:59) wrong based on 248 sessions
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Alejandro has a six-sided die with faces numbered 1 through 6. He rolls the die twice.
Quantity A |
Quantity B |
The probability that both rolls are even |
The probability that neither roll is a multiple of 3 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: Alejandro has a six-sided die with faces numbered 1 through
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21 Sep 2017, 04:39
21 Sep 2017, 04:39
6
The probability that the two rolls are even is 3/6*3/6 since there are three numbers out of six that are even (2, 4, 6). The probability that neither roll is a multiple of 3 is 4/6*4/6 because there are only two numbers who are multiple of 3, 3 and 6. Thus comparing 1/4 to 4/9 leads to the choice B!
General Discussion
Re: Alejandro has a six-sided die with faces numbered 1 through
[#permalink]
20 Dec 2018, 16:10
20 Dec 2018, 16:10
1
Answer: B
A: All possibilities: 6*6 = 36
The ones which both parameters are even:
([2 | 4 | 6] , 2) -> 3
([2 | 4 | 6] , 4) -> 3
([2 | 4 | 6] ,6) -> 3
Probability = 9/36
B: All possibilities: 6*6 = 36
The ones which neither roll is a multiple of 3:
We need to eliminate 3 and 6 from our numbers, so we will have: 4*4 = 16
Probability=16/36
B is bigger than A.
A: All possibilities: 6*6 = 36
The ones which both parameters are even:
([2 | 4 | 6] , 2) -> 3
([2 | 4 | 6] , 4) -> 3
([2 | 4 | 6] ,6) -> 3
Probability = 9/36
B: All possibilities: 6*6 = 36
The ones which neither roll is a multiple of 3:
We need to eliminate 3 and 6 from our numbers, so we will have: 4*4 = 16
Probability=16/36
B is bigger than A.
