Question

If three integers are chosen at random from the set of first 20 natural numbers. The chance that their product is multiple of 3 is

Answer 1

This problem is based on the probability. Here we are given the first 20 natural numbers. We will take the numbers which are multiple by 3 and the integers will be chosen at random. From that we need to find the probability that three numbers should be multiple of 3 and complete step by step explanation.

Formula used: $^n{C_r} = \dfrac{{n!}}{{(n - 1)!r!}}$
$n =$ Total number of objects in the set.
$r =$ The number of choosing objects from the set.

Complete step-by-step solution:
It is given that we have chosen three integers random from the set of first 20 natural numbers.
Hence from the given, we get,
$n = 20$ and $r = 3$
Substituting these values into combination formula mentioned in formula used,
Total number of ways ${ = ^{20}}{C_3}$
${ \Rightarrow ^{20}}{C_3} = \dfrac{{20!}}{{\left( {20 - 3} \right)!3!}}$
Subtracting the terms,
${ \Rightarrow ^{20}}{C_3} = \dfrac{{20!}}{{\left( {17} \right)!3!}}$
Solving the factorial,
${ \Rightarrow ^{20}}{C_3} = \dfrac{{20 \times 19 \times 18 \times 17!}}{{\left( {17} \right)!\left( {3 \times 2 \times 1} \right)}}$
Simplifying we get,
${ \Rightarrow ^{20}}{C_3} = 10 \times 19 \times 6$
Multiplying the terms we get,
${ \Rightarrow ^{20}}{C_3} = 1140$
Their product is multiple of 3 when at least one of the three numbers is multiple of 3.
Multiples of 3 up to first 20 natural numbers are
$3,6,9,12,15,18$
There are 6 nos.
Total favourable ways are by choosing any 1, 2 or 3 in these 6 numbers
So total favourable ways
Formulas mentioned in formula used, we get
$^6{c_1}{ \times ^{14}}{c_2}{ + ^6}{c_2}{ \times ^{14}}{c_1}{ + ^6}{c_3}{ \times ^{14}}{c_0}$
$\Rightarrow (6 \times 91) + (15 \times 14) + (5 \times 4)$
Multiplying the terms we get,
$\Rightarrow 546 + 210 + 20$
Adding we get,
$\Rightarrow 776$
Now the chance that their product is multiple of 3 is
Total favourable ways by total number of ways
That is $\dfrac{{776}}{{1140}}$
By simplifying, we get
Dividing numerator and denominator by 4, we get = $\dfrac{{194}}{{285}}$
$\therefore$ The probability of getting three integers are chosen at random from the set of first 20 natural numbers.

The chance that their product is multiple of 3 is $\dfrac{{194}}{{285}}$

Note: Permutations and combinations:
The various ways in which objects from a set may be selected, generally without replacement, to form subsets. This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.