Question

A number is chosen at random from the number $10$ to $99$. By seeing the number, a man will laugh if the product of the digits is $12$. If he chosen three numbers with replacement then the probability that he will laugh at least once is_______
A) \[1 - {(\dfrac{{43}}{{45}})^3}\]
B) \[1 - {(\dfrac{{43}}{{45}})^3}\]
C) $1 - {(\dfrac{{43}}{{44}})^3}$
D) $1 - {(\dfrac{{42}}{{43}})^3}$

Answer 1

Hint: Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes.

Complete step by step answer:
A number is chosen at random from $10$ to $99$.
$S = \{ 10,11,.....98,99\} $
Therefore, $n(s) = 90$
Let, A be an event that the man will laugh at if the product of the digits is $12$..
$A = \{ 26,34,43,62\} $
Let $P(x)$ denote the number of times the man laughs.
$\therefore P(x \ge 1) = 1 - P(x = 0)$
Now, $P(x = 0)$ implies that the three numbers are not from S.
Hence,
$P(x = 0)$ $ = {(\dfrac{{86}}{{90}})^3}$
$P(x = 0) = {(\dfrac{{43}}{{45}})^3}$
Hence, If the man chosen the three numbers with the replacement then the probability that he will laugh at least once is –
$P(x > 1) = 1 - {(\dfrac{{43}}{{45}})^3}$
This is the required answer.

Therefore from the given options, option A is the correct answer.

Note: The probability of any event always ranges between $0$ and $1$. It can never be negative nor the number greater than one.