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27 Apr 2022, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
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Question Stats:
55% (02:52) correct
45%
(02:42)
wrong
based on 181
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Two fair dice, each with sides numbered 1 to 6, are thrown simultaneously and their outcomes — the numbers on the sides facing up — are added. If it is assumed that the probability of getting x as the sum of the outcomes is equal to the probability of getting (x + 4) as the sum of the outcomes, then what is the probability of getting x as the sum of the outcomes?
A. 1/18
B. 1/12
C. 1/9
D. 5/36
E. 1/6
Are You Up For the Challenge: 700 Level Questions
A. 1/18
B. 1/12
C. 1/9
D. 5/36
E. 1/6
Are You Up For the Challenge: 700 Level Questions
27 Apr 2022, 05:54
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This is how I approach this question.
the sum of the outcomes can be 2, 3, 4, ..., 11, 12.
From the question it is asked, there is only one x, whose probability is the same as that for (x+4).
The probability of getting 2 and that of getting 12 are the same.
We have every reason to believe that the same pattern holds for 3 and 11, ..., and 5 and 9.
Because 9 = 5 +4.
we just need to figure out the probability of getting 5 as the sum of the outcome.
There are four combinations to get a sum of 5: 1 + 4, 2 + 3, 3 + 2, and 4 + 1.
For each one, the probability is 1/6 * 1/6 = 1/36
Altogether: 4 / 36 = 1/9
My answer is thus (C).
the sum of the outcomes can be 2, 3, 4, ..., 11, 12.
From the question it is asked, there is only one x, whose probability is the same as that for (x+4).
The probability of getting 2 and that of getting 12 are the same.
We have every reason to believe that the same pattern holds for 3 and 11, ..., and 5 and 9.
Because 9 = 5 +4.
we just need to figure out the probability of getting 5 as the sum of the outcome.
There are four combinations to get a sum of 5: 1 + 4, 2 + 3, 3 + 2, and 4 + 1.
For each one, the probability is 1/6 * 1/6 = 1/36
Altogether: 4 / 36 = 1/9
My answer is thus (C).
14 Nov 2024, 08:32
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sristiiiiiiiii
Two fair dice, each with sides numbered 1 to 6, are thrown simultaneously and their outcomes — the numbers on the sides facing up — are added. If it is assumed that the probability of getting x as the sum of the outcomes is equal to the probability of getting (x + 4) as the sum of the outcomes, then what is the probability of getting x as the sum of the outcomes?
A. 1/18
B. 1/12
C. 1/9
D. 5/36
E. 1/6
We are told that the probability of getting x as the sum of the outcomes when rolling two fair dice is equal to the probability of getting the sum x + 4, for some specific number x.
• Can x be 2? That would imply that the probability of getting sums of 2, 6, and 10 is equal. This is not correct because we can get 2 only in 1 way ((1, 1)), while we can get 6 and 10 in more ways.
• Can x be 3? This would mean that the probability of getting sums of 3, 7, and 11 is equal, which is also incorrect. We can get 3 only in 2 ways ((1, 2) or (2, 1)), but we can get 7 in more ways: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), and (4, 3).
• Can x be 4? This would imply that the probability of getting sums of 4, 8, and 12 is equal, which is also incorrect. We can get 12 only in 1 way ((6, 6)), while we can get 4 and 8 in more ways.
• Can x be 3? This would mean that the probability of getting sums of 3, 7, and 11 is equal, which is also incorrect. We can get 3 only in 2 ways ((1, 2) or (2, 1)), but we can get 7 in more ways: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), and (4, 3).
• Can x be 4? This would imply that the probability of getting sums of 4, 8, and 12 is equal, which is also incorrect. We can get 12 only in 1 way ((6, 6)), while we can get 4 and 8 in more ways.
Thus, x must be 5, and x + 4 would be 9 (the only two remaining numbers). Indeed, both can be obtained in the same number of ways. We can get the sum of 5 as (1, 4), (4, 1), (2, 3), and (3, 2), so in four ways. Similarly, we can get 9 in four ways: (3, 6), (6, 3), (4, 5), and (5, 4).
Therefore, the probability of getting x = 5 is 4/36 = 1/9.
Answer: C.
