Question

A number is chosen at random from the numbers $10$ to $99$. By seeing the number a man will laugh if product of the digits is $12$. If he choose three numbers with replacement then the probability that he will laugh at least once is

• A. $1-{\left(\frac{3}{5}\right)}^3$
• B. ${\left(\frac{43}{45}\right)}^3$
• C. $1-{\left(\frac{4}{25}\right)}^3$
• D. $1-{\left(\frac{43}{45}\right)}^3$

Answer 1

The correct option is D $1-{\left(\frac{43}{45}\right)}^3$

There can be four such numbers, i.e., $43,34,62,26$

whose product of digit is $12$.

Hence, Probability that a man will laugh by seeing the chosen numbers = $\frac{4}{90}=\frac{2}{45}$

$\therefore ,$ the required probability = $1-{(1-\frac{2}{45})}^3=1-{\left(\frac{43}{45}\right)}^3$