Question

Two numbers are selected at random from 1, 2, 3, ….., 100 and are multiplied, then probability (correct to two places of decimals) that the product thus obtained is divisible by 3 is
$\left( a \right)0.22$
$\left( b \right)0.33$
$\left( c \right)0.44$
$\left( d \right)0.55$

Answer 1

In this particular question use the concept that if there are n different objects and we have choose r objects from them so the number of ways of choosing are given as ${}^n{C_r}$ and use the concept that probability is the ratio of favorable number of outcomes to the total number of outcomes so use these concepts to reach the solution of the question.

Complete step by step answer:
Given data:
Numbers available are
1, 2, 3, 4, ....., 100
Now we have to select two numbers and are multiplied, then probability (correct to two places of decimals) that the product thus obtained is divisible by 3.
So first find out the numbers which are divisible by 3.
So the set of numbers are, 3, 6, 9, 12,....... 99
So as we see that the above series forms an A.P, with first term, a = 3, common difference, d = (6 – 3) = (9 – 6) = 3, last term, ${a_n} = 99$
Let the number of terms is n.
Now as we know that formula for the last term of an A.P is given as,
$\Rightarrow {a_n} = a + \left( {n - 1} \right)d$, where symbols have their usual meaning.
$\Rightarrow 99 = 3 + \left( {n - 1} \right)3$
$\Rightarrow n - 1 = \dfrac{{99 - 3}}{3}$
$\Rightarrow n - 1 = 32$
$\Rightarrow n = 33$
So there are 33 terms available which are divisible by 3.
So the remaining numbers are 100 – 33 = 67.
Now as we know that if there are n different objects and we have choose r objects from them so the number of ways of choosing are given as ${}^n{C_r}$.
So, the number of ways to select 2 numbers out of 100 are ${}^{100}{C_2}$.
So the total number of outcomes = ${}^{100}{C_2}$.
Now the favorable cases are given below:
Case – $\left( 1 \right)$ When both of the numbers are chosen by the numbers which are multiple of 3.
As there are 33 numbers available which are multiple of 3.
So, the number of ways to select 2 numbers out of 33 are ${}^{33}{C_2}$.
Case – $\left( 2 \right)$ When one number is choose from the multiple of 3 and other number is chosen from the remaining numbers,
So the number of ways to choose one number out of 33 are ${}^{33}{C_1}$
And the number of ways to choose one number out of the remaining numbers i.e. 67 are ${}^{67}{C_1}$
So the total number of ways to one number is choose from the multiple of 3 and other number is chosen from the remaining numbers is the product of above two calculated values = ${}^{33}{C_1} \times {}^{67}{C_1}$
So the total number of favorable cases are the sum of the above two cases.
So the total number of favorable cases are = ${}^{33}{C_2} + {}^{33}{C_1} \times {}^{67}{C_1}$
Now as we know that probability is the ratio of favorable number of outcomes to the total number of outcomes.
So, ${\text{P}} = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}$
$\Rightarrow {\text{P}} = \dfrac{{{}^{33}{C_2} + {}^{33}{C_1} \times {}^{67}{C_1}}}{{{}^{100}{C_2}}}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property we have,
$\Rightarrow {\text{P}} = \dfrac{{\dfrac{{33!}}{{2!\left( {33 - 2} \right)!}} + \dfrac{{33!}}{{1!\left( {33 - 1} \right)!}} \times \dfrac{{67!}}{{1!\left( {67 - 1} \right)!}}}}{{\dfrac{{100!}}{{2!\left( {100 - 2} \right)!}}}}$
Now simplify this we have,
$\Rightarrow {\text{P}} = \dfrac{{\dfrac{{33!}}{{2!\left( {31} \right)!}} + \dfrac{{33!}}{{1!\left( {32} \right)!}} \times \dfrac{{67!}}{{1!\left( {66} \right)!}}}}{{\dfrac{{100!}}{{2!\left( {98} \right)!}}}}$
$\Rightarrow {\text{P}} = \dfrac{{\dfrac{{33.32.31!}}{{\left( {2.1} \right)\left( {31} \right)!}} + \dfrac{{33.32!}}{{\left( {32} \right)!}} \times \dfrac{{67.66!}}{{\left( {66} \right)!}}}}{{\dfrac{{100.99.98!}}{{\left( {2.1} \right)\left( {98} \right)!}}}}$
$\Rightarrow {\text{P}} = \dfrac{{\left( {33 \times 16} \right) + \left( {33 \times 67} \right)}}{{\left( {50 \times 99} \right)}}$
$\Rightarrow {\text{P}} = \dfrac{{16 + 67}}{{50 \times 3}} = \dfrac{{83}}{{150}} = 0.55$
So this is the required probability.

So, the correct answer is “Option D”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of combination as well as probability which is stated above, so first construct the favorable number of outcomes and total number of outcomes as above calculated then divide them as above and simplify using formula of combination as above we will get the required probability.