Source: Magoosh
One number, k, is selected at random from a set of 11 consecutive even integers. What is the probability that k = 10?
1) The average (arithmetic mean) of the set is zero.
2) The probability that k = 10 is the same as the probability that k = -10.
The OA is A
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One number, k, is selected at random from a set of 11
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BTGmoderatorLU
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Jay@ManhattanReview
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If k = 10 is a member of the set, then the probability that the number randomly selected in 10 = 1/11; however,BTGmoderatorLU wrote:Source: Magoosh
One number, k, is selected at random from a set of 11 consecutive even integers. What is the probability that k = 10?
1) The average (arithmetic mean) of the set is zero.
2) The probability that k = 10 is the same as the probability that k = -10.
The OA is A
If k = 10 is a NOT member of the set, then the probability that the number randomly selected in 10 = 0.
So, we have to determine whether 10 is a member of the set.
Let's take each statement one by one.
1) The average (arithmetic mean) of the set is zero.
A set such that it has 11 consecutive numbers in it, and its mean is 0 is: {-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}.
The probability that the number randomly selected in 10 = 1/11. Sufficient.
2) The probability that k = 10 is the same as the probability that k = -10.
Case 1: Say set is: {-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}.
The probability that the number randomly selected in 10 = 1/11.
Case 2: Say set is: {12, 14, 16, 18, 20, 22, 24, 26, 28,30, 32}.
The probability that the number randomly selected in 10 = 0.
No unique answer. Insufficient.
The correct answer: A
Hope this helps!
-Jay
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deloitte247
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What is the probability that k=10
Statement 1
The average arithemetic mean of the set is zero bearing in mind that the set of 11 consecutive integers every elements in the set = n+2, the set will definitely start from -10
$$Set=\left\{-10,-8,-6,-4,-2,\ 0,\ 2,\ 4,\ 6,\ 8,\ 10\right\}$$
Probability that k=10
$$_{=\ }\frac{no\ of\ occurence}{Total\ no\ of\ event}=\frac{1}{11}$$
Statement 1 is INSUFFICIENT.
Statement 2
The probability that k=10
is same as the probability that k=-10 both 10 and -10 cannot be in any set of 11 consecutive numbers except for the one in statement 1, so it is either that probability is 1 or we have a set that is neither inside , in this case probability is 0. Hence statement 2 is INSUFFICIENT.
Statement 1 alone is SUFFICIENT.
$$answer\ is\ Option\ A$$
Statement 1
The average arithemetic mean of the set is zero bearing in mind that the set of 11 consecutive integers every elements in the set = n+2, the set will definitely start from -10
$$Set=\left\{-10,-8,-6,-4,-2,\ 0,\ 2,\ 4,\ 6,\ 8,\ 10\right\}$$
Probability that k=10
$$_{=\ }\frac{no\ of\ occurence}{Total\ no\ of\ event}=\frac{1}{11}$$
Statement 1 is INSUFFICIENT.
Statement 2
The probability that k=10
is same as the probability that k=-10 both 10 and -10 cannot be in any set of 11 consecutive numbers except for the one in statement 1, so it is either that probability is 1 or we have a set that is neither inside , in this case probability is 0. Hence statement 2 is INSUFFICIENT.
Statement 1 alone is SUFFICIENT.
$$answer\ is\ Option\ A$$
