Question

If one? three-digit number? (0 cannot be a left? digit) is chosen at random from all those that can be made from the following set of? digits, find the probability that the one chosen is not a multiple of 5 . (0,1,2,3,4,5,6)

Answer 1

Solution

Since the question does not explicitly say that each digit can occur only once in any 3-digit number formed, we will assume that any of the given 7 digits can appear any number of times in any 3-digit number formed. This is natural because 111, 232, 445, etc are all valid numbers.

Back-up Theory

Probability of an event E, denoted by P(E) = n/N …………………………(1)

where n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and N = n(S) = Total number all possible outcomes/cases/possibilities.

Number of k-digit numbers that can be formed with n distinct digits (k ? n)

= ⁿP_k = (n!)/(n - k)! if repetition is not permitted and digit 0 is not included in the n digits ..(2)

= n^k, if repetition is permitted and digit 0 is not included in the n digits ………………….(3)

= (n - 1)[(n - 1)!/{(n- 1) – (k - 1)}!], if repetition is not permitted and digit 0 is one of the n digits ……………………………………………………………………………………….(4)

= (n - 1)n^{(k - 1)}, if repetition is permitted and digit 0 is one of the n digits ………………..(5)

A number is a multiple of 5 if the last digit (i.e., units digit) is 0 or 5. ……………………..(6)

Now, to work out the answer,

With repetition permitted

Given 7 digits, (0,1,2,3,4,5,6), number of 3-digit that can be formed using these digits,

vide (5), is (7 - 1)7^{(3 - 1)}

= 6 x 49

= 294

Number of 3-digit which are multiples of 5 that can be formed using these digits = (6 x 7) + (6 x 7) = 84

Thus, the required probability. Vide (1), is 84/294

= 0.2857 ANSWER