Case i: Set A = {-17, -16, -15, -14, -13, -12,. . . . . . -2, -1, 0, 1, 2, . . . . , 13, 14, 15, 16, 17}
Since 0 is the median, we get:
QUANTITY A: Probability of selecting an even number less than the median from set A = 8/35
QUANTITY B: Probability of selecting an odd number greater than the median from set A = 9/35
In this case, Quantity B is greater

Case ii: Set A = {-16, -15, -14, -13, -12,. . . . . . -2, -1, 0, 1, 2, . . . . , 13, 14, 15, 16, 17, 18}
Since 1 is the median, we get:
QUANTITY A: Probability of selecting an even number less than the median from set A = 9/35
QUANTITY B: Probability of selecting an odd number greater than the median from set A = 8/35
In this case, Quantity A is greater

Answer: D

*** If a set contains consecutive nos. starting with odd and ending with even no. eg - 1 ,2 , .... 100 then we have 50 odd and 50 even


Now the median will be between 17th and 18th no. so there will be seventeen no. below the median and 18 no. above the median.


Now if the set start with odd no. then there will be 9 odd and 8 even number so the probability of selecting an even no below median = \(\frac{8}{35}\)

and the probability of selecting odd no. above median will be = \(\frac{9}{35}\)




Similarly


If the set start with even no. then the probability of selecting even no. below mean will be = \(\frac{9}{35}\)

and the probability of selecting odd no. above mean = \(\frac{8}{35}\)

Hence option is D

May be this is not ideal way bit i created my own set of numbers with n=5.
1,2,3,4,5
2,3,4,5,6
Based on the explanations above, we can deduce the same answer without dealing with 35 numbers

Greenlight has the more understanble answer. But, keep in mind that:

(1) Since you don´t know if the number starting on even or odd, the answer is D. Why?

(2) Depends on even or odd number, there are different probabilities.

If it´s starting on even, then A = 9/35

If it´s starting on odd, then A = 8/35

Therefore, how can I know? Clearly D.

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