Question

# A number is chosen at random from the numbers 10 to 99 by seeing that the number, a man will laugh if the product of the digits is 12. If he has chosen three numbers with a replacement then the probability that he will laugh at least once is(A) $1-{{\left( \dfrac{43}{45} \right)}^{3}}$(B) $1-{{\left( \dfrac{2}{45} \right)}^{3}}$(C) ${{\left( \dfrac{43}{45} \right)}^{3}}$(D) ${{\left( \dfrac{2}{45} \right)}^{3}}$

Hint: We have to choose any number from 10 to 99. The total numbers from 10 to 99 is equal to 90. Since we are choosing any number from these 90 numbers, the sample space for this experiment is equal to 90. The two digits numbers for which the product of the digits is equal to 12 are 26,34,43 and 62. Now, get the probability of laughing the man when we draw a number using the formula, $\text{Probability=}\dfrac{\text{Number}\,\text{of}\,\text{favorable}\,\text{outcomes}}{\text{Sample}\,\text{Space}}$ . Now, deduct this probability from 1 and get the probability that the man doesn’t laugh while drawing a number. For the whole event, we can say that $P\left( x=0 \right)+P\left( x\ge 1 \right)=1$ . Here, $P\left( x=0 \right)$ is the probability that the man doesn't laugh while drawing three numbers and $P\left( x\ge 1 \right)$ is the probability that the man laughs at least once while drawing three numbers. Using the probability that the man doesn’t laugh while drawing a number, calculate the probability that the man doesn't laugh while drawing three numbers, $P\left( x=0 \right)$ . Now, put the value of $P\left( x=0 \right)$ in the equation $P\left( x=0 \right)+P\left( x\ge 1 \right)=1$ and get the value of $P\left( x\ge 1 \right)$ .

Complete step by step solution:
According to the question, we have been given that a number is chosen at random from the numbers 10 to 99 by seeing that the number, a man will laugh if the product of the digits is 12.
We can calculate the total numbers from 10 to 99.
The total numbers from 10 to 99 = 90.
Since we are choosing any number from these 90 numbers, the sample space for this experiment is equal to 90.
The sample space = 90 ……………………………………(1)
All the numbers that are to be chosen from the numbers which are lying between 10 and 99 are two digits numbers. So, we have to first figure out the possible two-digit numbers for which the product of the digits is equal to 12.
The two digits numbers for which the product of the digits is equal to 12 are 26,34,43 and 62.
We have four possible numbers for which the product of the digits is equal to 12. So, to make the man laugh we have to choose any number from these four numbers.
The number of favorable outcomes to make that man laugh = 4 ……………………………..(2)
We know the formula, $\text{Probability=}\dfrac{\text{Number}\,\text{of}\,\text{favorable}\,\text{outcomes}}{\text{Sample}\,\text{Space}}$ ……………………………………..(3)
Now, from equation (1), equation (2), and equation (3), we get
The probability of laughing the man when we draw a number = $\dfrac{4}{90}$ = $\dfrac{2}{45}$ …………………………………(4)
The probability that the man doesn’t laugh while drawing a number= $1-\dfrac{2}{45}$ = $\dfrac{43}{45}$ ………………………………..(5)
For the whole event, we can say that
$P\left( x=0 \right)+P\left( x\ge 1 \right)=1$ ………………………………….(6)
Here, $P\left( x=0 \right)$ is the probability that the man doesn't laugh while drawing three numbers and $P\left( x\ge 1 \right)$ is the probability that the man laughs at least once while drawing three numbers.
If the man doesn't laugh while drawing three numbers. It means that any numbers are drawn from 10 to 99 except 26, 34, 43, and 62.
From equation (5), we have the probability that the man doesn’t laugh while drawing a number
The probability that the man doesn't laugh while drawing three numbers, $P\left( x=0 \right)$ = $\left( \dfrac{43}{45} \right)\times \left( \dfrac{43}{45} \right)\times \left( \dfrac{43}{45} \right)={{\left( \dfrac{43}{45} \right)}^{3}}$ ………………………………………..(7)
Now, from equation (6) and equation (7), we get
\begin{align} & \Rightarrow P\left( x=0 \right)+P\left( x\ge 1 \right)=1 \\ & \Rightarrow {{\left( \dfrac{43}{45} \right)}^{3}}+P\left( x\ge 1 \right)=1 \\ \end{align}
$\Rightarrow P\left( x\ge 1 \right)=1-{{\left( \dfrac{43}{45} \right)}^{3}}$ ………………………………….(8)
Therefore, the probability that the man laughs at least once while drawing three numbers is $1-{{\left( \dfrac{43}{45} \right)}^{3}}$ .
Hence, the correct option is (A).

Note: In this question, one might take the probability that the man doesn't laugh while drawing three numbers equal to $\left( \dfrac{43}{45} \right)$ and then deduct it from 1 to get the probability that the man laughs at least once while drawing three numbers. This is wrong. Since we have to choose three numbers, we have to multiply the probability that the man doesn't laugh while a number thrice. After multiplying we get the probability that the man doesn't laugh while drawing three numbers equal to $\left( \dfrac{43}{45} \right)\times \left( \dfrac{43}{45} \right)\times \left( \dfrac{43}{45} \right)={{\left( \dfrac{43}{45} \right)}^{3}}$ .