alex1233 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
I would use one of the approaches already suggested.
That said, below is another approach -- one that might prove useful for other probability problems.
good cases = all possible cases - bad casesLet Group A represent the 4 values above 27:
28, 29, 30, 31
Let Group B represent the 4 values below 27:
23, 24. 25, 26
All possible cases:From 8 integers, the number of ways to select 4 = 8C4 \(= \frac{8*7*6*5}{4*3*2*1} = 70\)
Bad case 1: 4 integers are selected from Group A or from Group B, with the result that 27 will not be the median of the resulting 5 numbersNumber of options for the selected group = 2 (A or B)
Number of ways to select 4 integers from this group = 4C4 \(= \frac{4*3*2*1}{4*3*2*1} = 1\)
To combine these options, we multiply:
2*1 = 2
Bad case 2: 3 integers are selected from Group A or from Group B, with the result that 27 will not be the median of the resulting 5 numbersNumber of options for the group from which 3 integers are selected = 2 (A or B)
Number of ways to select 3 integers from this group = 4C3 \(= \frac{4*3*2}{3*2*1} = 4\)
Number of ways to select 1 integer from the other group = 4C1 = 4
To combine these options, we multiply:
2*4*4 = 32
Good cases:all possible cases - bad case 1 - bad case 2 = 70 - 2 - 32 = 36
Probability that 27 is the median:\(\frac{good-cases}{all-possible-cases }= \frac{36}{70} = \frac{18}{35}\)