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 chooses three numbers with replacement then the probability that he will laugh atleast 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

Answer: (D)

$1 - \left( \frac { 43 } { 45 } \right) ^ { 3 }$

Answer 2

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

whose product of digit is 12.

Ž Probability that the man will laugh by seeing the chosen number = $\frac { 4 } { 90 } = \frac { 2 } { 45 }$.

Ž Required probability = 1 - $\left( 1 - \frac { 2 } { 45 } \right) ^ { 3 } = 1 - \left( \frac { 43 } { 45 } \right) ^ { 3 }$.