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 chosen three numbers with replacement then the probability that he will laugh at least once is ............

• A. $1-{\left(\frac{43}{45}\right)}^3$
• B. $1-{\left(\frac{2}{45}\right)}^3$
• C. ${\left(\frac{43}{45}\right)}^3$
• D. ${\left(\frac{2}{45}\right)}^3$

Answer 1

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

The numbers between 10-99 whose product of the digits give 12 are
$26,34,43,62-S$
Let $P\left(X\right)$ denote the number of times the man laughs.
$P(X\ge 1)$
$=1-P(X=0)$
Now P(x=0) implies that all the three numbers are not from S
Hence
$P(X=0)$
$=(\frac{86}{90}{)}^3$
$=(\frac{43}{45}{)}^3$
Hence
$P(X\ge 1)$
$=1-(\frac{43}{45}{)}^3$