Question

Two numbers are selected at random from $1,2,3,4,....100$ and multiplied. The probability that the product thus obtained is divisible by $3$ is.........
$({\text{A) 0}}{\text{.55}}$
$({\text{B) 0}}{\text{.62}}$
$({\text{C) 0}}{\text{.49}}$
$({\text{D) 0}}{\text{.60}}$

Answer 1

Here, we have to find the probability that the product of two numbers from a given number is divisible by $3$.
First, we need to find the number of cases of getting the number divisible by$3$.
Then, we have to divide this by the total number of cases of getting two random numbers from the number.
Finally we get the required answer.

Formula used: ${}^{_n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}}$

Complete step-by-step solution:
Total number of cases obtained by multiplying two numbers out of $100 = {}^{100}{{\text{C}}_{\text{2}}}$
Let us say the two numbers are ${\text{x}}$ and ${\text{y}}$.
Then, ${\text{x }} \times {\text{ y}}$ = divisible by $3$ or$3{\text{c}}$, where ${\text{c}}$ is any constant number between $1$ and$100$.
So, there are two cases by which we get the product divisible by $3$
Case one: Any one of the numbers is divisible by $3$
Case two: Both the numbers are divisible by $3$.
We will get a product divisible by $3$ in both the cases.
Case 1:
Out of $100$ numbers, the numbers divisible by $3$ are $3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99$ which are $33$ numbers in total.
Here, ${\text{x}}$ will be any one of the numbers among these $33$ numbers.
Since the number is one among $33$ numbers, it can be expressed as ${}^{33}{{\text{C}}_1}$
Ultimately, the other number will be a number which is not in these $33$ numbers, i.e., among the remaining $67$ numbers in $1,2,3,4,....100$
Since the other number is one among $67$ numbers, it can be expressed as ${}^{67}{{\text{C}}_1}$
Therefore, Number of cases of getting one number divisible by $3$and other is non-divisible of $3$ = $\dfrac{{{}^{33}{{\text{C}}_1}\; \times {\text{ }}{}^{67}{{\text{C}}_1}}}{{{}^{100}{{\text{C}}_2}}}....\left( 1 \right)$
Here, we divide ${}^{100}{{\text{C}}_2}$ because the required two numbers is been selected from the total number of cases, ${}^{100}{{\text{C}}_2}$.
Case two:
Both the numbers are divisible by $3$.
We know that, there are $33$ numbers which are divisible by $3$ in $1,2,3,4,....100$.
Therefore, numbers of ways of getting two numbers out of $33$ numbers is ${}^{33}{{\text{C}}_2}$
Number of ways of getting both the numbers divisible by $3$ among total number of cases =$\dfrac{{{}^{33}{{\text{C}}_2}}}{{{}^{100}{{\text{C}}_2}}}....\left( 2 \right)$.
Let us adding $\left( 1 \right)$and $\left( 2 \right)$ we get
The total number of cases of getting a number divisible by $3$ =$\dfrac{{{}^{33}{{\text{C}}_1}\; \times {\text{ }}{}^{67}{{\text{C}}_1}}}{{{}^{100}{{\text{C}}_2}}} + \dfrac{{{}^{33}{{\text{C}}_2}}}{{{}^{100}{{\text{C}}_2}}}$
On rewriting the terms we get
=$\dfrac{{\left( {{}^{33}{{\text{C}}_1}\; \times {\text{ }}{}^{67}{{\text{C}}_1}} \right)\;{\text{ + }}{}^{33}{{\text{C}}_2}}}{{{}^{100}{{\text{C}}_2}}}$
Expanding the terms by using the formula and we get
$= \dfrac{{\left( {\dfrac{{33!}}{{1!(33 - 1)!}}\; \times {\text{ }}\dfrac{{67!}}{{1!(67 - 1)!}}} \right)\;{\text{ + }}\dfrac{{33!}}{{2!(33 - 2)!}}}}{{\dfrac{{100!}}{{2!(100 - 2)!}}}}$
On subtracting the denominator terms and we get
$= \dfrac{{\left( {\dfrac{{33 \times 32!}}{{(32)!}}\; \times {\text{ }}\dfrac{{67 \times 66!}}{{(66)!}}\;} \right){\text{ + }}\dfrac{{33 \times 32 \times 31!}}{{2 \times 1(31)!}}}}{{\dfrac{{100 \times 99 \times 98!}}{{2 \times 1(98)!}}}}$
On cancelling the same term and we get
$= \dfrac{{\left( {33\; \times {\text{ 67}}} \right)\;{\text{ + }}\dfrac{{33 \times 32}}{2}}}{{\dfrac{{100 \times 99}}{2}}}$
Let us multiply we get,
$= \dfrac{{2211\;{\text{ + }}\dfrac{{1056}}{2}}}{{\dfrac{{9900}}{2}}}$
On dividing the terms and we get
$= \dfrac{{2211\;{\text{ + 528}}}}{{4950}}$
Let us add the numerator term and we get
$= \dfrac{{2739}}{{4950}}$
On dividing the terms and we get
$= 0.5533$
Rewrite the terms and we get
$= 0.55$

Therefore $({\text{A) 0}}{\text{.55}}$ is the correct answer.

Note: These types of questions from probability are solved either using combination or permutation according to the nature of the question.
A combination is used to determine the number of possible arrangements in the given set of collection of items where the order of the selection does not matter. In probability it is used to find the number of ways a number can be drawn or taken out of the given set. Whereas,
Permutation is used to find the number of ways to arrange the given set in a sequence or linear order. In probability, it is used to find the number of cases the selected numbers or set can be arranged in a sequence.