Question

Assume that 3 digits are selected at random from the set {1,3,4,5,6,7}{1,3,4,5,6,7} and are arranged in random order.

What is the probability that the resulting 3-digit number is less than 600?

Answer 1

When arranging numbers in any order, the order is important. For eg. if 3,4,6 are selected then number 346 is not same as 436.

Therefore here for calculating number of ways to choose 3 nos. from set of 6 nos. will be given by permutation.

Probability= no. of desired outcomes / Total outcomes.

Here we assume repetition is not allowed so we can not have numbers like 333,445,etc.

Total

There are 3 places so if for the first place we have 6 options

then for the second place we will have 5 options since 1 is already chosen

Then for the 3rd place we will have 4 options since 2 numbers are already chosen.

So number number of ways = 6 * 5 * 4

= 120

OR

Permutation = 3 out of 6

= 6P3

= 6!/ (6-3)! ...............

= 120

Desired

We need a number less than 600.

So we can make sure that the first digit should be less than 5. Therefore for the first place we can either have 1,3,4,5 that is 4 options.

But for the second place we can have any number. So there are 5 options (1 already chosen for first place).

For the third place now we 4 options left since 2 numbers are chosne

Desired outcomes = 4 * 5 * 4

= 80

OR

for first place we have to choose 1 from 4 option = 4P1 = 4

For next two we need to choose two from remaining 5 = 5P2 = 20

Prob = 80 / 120

Ans: 0.667