BTGmoderatorLU wrote:There are 8 disks in a container that are numbered 23, 24, 25, 26, 28, 29, 30 and 31. 4 disks are randomly removed from the container, one after the other, without replacement. Then a list of 5 numbers containing the 4 numbers on the disks selected, together with the number 27 will be made. What is the probability that 27 is the median of the list of 5 numbers?
A. 2/5
B. 17/35
C. 1/2
D. 18/35
E. 5/9
P = (good outcomes)/(all possible outcomes).
All possible outcomes:
From the 8 numbers, the number of ways to choose 4 = 8C4 = (8*7*6*5)/(4*3*2*1) = 70.
Good outcomes:
For 27 to be the median of the 5 numbers, two of the 4 numbers selected must be LESS THAN 27, while the other two must be GREATER THAN 27.
Of the 8 numerical options, four are less than 27 (23, 24, 25, 26), and four are greater than 27 (28, 29, 30, 31).
From the 4 numbers less than 27, the number of ways to choose 2 to be below the desired median = 4C2 = (4*3)/(2*1) =
6.
From the 4 numbers greater than 27, the number of ways to choose 2 to be above the desired median = 4C2 = (4*3)/(2*1) =
6.
To combine the options in blue, we multiply:
6*6 = 36.
Thus:
P(27 is the median) = 36/70 = 18/35.
The correct answer is
D.
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