Question

Three different numbers are selected at random from the set $A=\left\{ 1,2,3,...,10 \right\}$. The probability that the product of two of the numbers is equal to third is:
A. $\dfrac{3}{4}$
B. $\dfrac{1}{40}$
C. $\dfrac{1}{8}$
D. none of these

Answer 1

Hint: Here we will first calculate the number of elements in the sample space which is equal to the number of elements in the given set $A$. Here we need to calculate the probability for the event that when the three elements are selected at random from the set, that the product of two of the numbers is equal to the third. So here we will look into the given set and list out all the elements that are present in the set $A$, which will satisfy the given condition. From this we will calculate the probability by taking the ratio of number of possible outcomes to the total number of outcomes for the event.

Complete step-by-step answer:
Given that, set $A$ is
$A=\left\{ 1,2,3,...,10 \right\}$
Number of elements in the set $A$ is $10$.
When we multiply the given set with number One, then we will get
$1.A=\left\{ \text{1,2,3,4,5,6,7,8,9,10} \right\}$
When we multiply the given set with number Two, then we will get
$2.A=\left\{ \text{2,4,6,8,10,12,14,16,18,20} \right\}$
When we multiply the given set with number Three, then we will get
$3.A=\left\{ \text{3,6,9,12,15,18,21,24,27,30} \right\}$
When we multiply the given set with number Four, then we will get
$4.A=\left\{ \text{4,8,12,16,20,24,28,32,36,40} \right\}$
When we multiply the given set with number Five, then we will get
$5.A=\left\{ \text{5,10,15,20,25,30,35,40,45,50} \right\}$
When we multiply the given set with number Six, then we will get
$6.A=\left\{ \text{6,12,18,24,30,36,42,48,54,60} \right\}$
When we multiply the given set with number Seven, then we will get
$7.A=\left\{ \text{7,14,21,28,35,42,49,56,63,70} \right\}$
When we multiply the given set with number Eight, then we will get
$8.A=\left\{ \text{8,16,24,32,40,48,56,64,72,80} \right\}$
When we multiply the given set with number Nine, then we will get
$9.A=\left\{ \text{9,18,27,36,45,54,63,72,81,90} \right\}$
When we multiply the given set with number Ten, then we will get
$10.A=\left\{ \text{10,20,30,40,50,60,70,80,90,100} \right\}$
Here we can observe that when we pick a random two numbers from the set $A$, then their product also in set $A$, for below numbers
$\begin{align}
  & 2\times 3=6 \\
 & 2\times 4=8 \\
 & 2\times 5=10 \\
\end{align}$
Hence the number of possible cases for the event is $3$.
The number of ways to select Three numbers from a set of Ten numbers is given by ${}^{10}{{\text{C}}_{3}}$.
Hence the required probability is
$\begin{align}
  & P=\dfrac{3}{{}^{10}{{\text{C}}_{3}}} \\
 & =\dfrac{3}{\dfrac{10!}{3!\left( 10-3 \right)!}} \\
 & =\dfrac{3}{\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}} \\
 & =\dfrac{1}{40} \\
\end{align}$
Hence the probability of selecting three numbers randomly from set $A$, that the product of two is equal to third is $\dfrac{1}{40}$

So, the correct answer is “Option (b)”.

Note: While observing the multiples of Two numbers in the set, students may write
$\begin{align}
  & 3\times 2=6 \\
 & 4\times 2=8 \\
 & 5\times 2=10 \\
\end{align}$
And notes the number of favorable outcomes as $3+3=6$. But it doesn’t give the right answer, because multiplication of numbers obeys associative law hence $2\times 3=3\times 2=6$etc. so there is no need to count all those things.