Question

There are ninety cards in a box numbered $10,11,12,\cdots ,98,99$. Three cards are drawn from the box one by one with replacement. The probability that product of the digits on the cards will be $12$ at least once is

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

Answer 1

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

Only four cards are there, product of whose digits is $12$.

These are $34,43,62$ and $26$.
The probability such that we do not get the required product in one card picked $=\frac{(90-4)}{90}=\frac{43}{45}$
The probability that we do not get the required product in none of the three draws $={\left(\frac{43}{45}\right)}^3$

There, the final answer is $1-{\left(\frac{43}{45}\right)}^3$.

Hence, option C.