Bunuel wrote:

3 numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. If the first number selected is w, the second number selected is x, and the third number is y, what is the probability that w < x < y ?

A. 3/40

B. 28/243

C. 3/25

D. 33/100

E. 64/125

Kudos for a correct solution.

Total number of ways to select 3 numbers from the given set will be 10 x 10 x 10 = 1000

The number of combinations in which w < x < y will be

1) when w = 0

let w= 0 and x=1 then y can have 8 values let w= 0 and x=2 then y can have 7 values

This pattern will continue till y can only 1 value i.e 9

So for w=0 we will have (8+7+6+5+4+3+2+1) combinations

or 8(9)/2 =

36 ... ( sum of first n natural number is given by n(n+1)/2 )

2) when w=1

let w=1 and x=2 then y can have 7 values let w=1 and x=3 then y can have 6 values

This pattern will again continue till y can only 1 value i.e 9

number of combinations in this case will be 7(8)/2

=

28Notice that for each increasing value of w, the maximum number of values which y can have, decreases by 1. This can be seen in highlighted statements.

So, we just have to subtract the maximum value of y from the sum in each case to get the sum of next case. This will save time as we do not have to calculate the number of combinations for every case.

So the total number of combinations will be

36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

So the probability will be

120/1000

= 3/25

Answer:- C