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 atleast once is

• A. $1-{\left(\frac{31}{45}\right)}^3$
• B. $1-{\left(\frac{43}{45}\right)}^3$
• C. $1-{\left(\frac{42}{43}\right)}^3$
• D. $1-{\left(\frac{41}{45}\right)}^3$

Answer 1

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

Given numbers are 10, 11, 12, 13, . . . , 99. The numbers with their digits product is 12 are 26, 34, 43, 62 ie, only 4.
$\therefore P=P\left(E\right)=\frac{4}{90}=\frac{2}{45}q=P\left(\overline{E}\right)=1-\frac{2}{45}=\frac{43}{45}$
Hence, the probability that he will laugh at least once
$=1-q^3=1-{\left(\frac{43}{45}\right)}^3$