Question

One ticket is selected at random from $50$ tickets numbered $00,01,02,...,49$. Then the probability that the sum of the digits on the selected ticket is $8$, given that the product of these digits is zero, equals

• A. $1/14$
• B. $1/7$
• C. $5/14$
• D. $1/50$

Answer 1

The correct option is A $1/14$

A $=$Events that sum of the digits on selected ticket is 8
$=\{08,17,26,35,44\}$

Therefore, total $5$ cases when sum of digits is $8$

$\Rightarrow n\left(A\right)=5$

Event that product of digits is zero

$=$ $\{00,01,02,03,04,05,06,07,08,09,10,20,30,40\}$

$\Rightarrow n\left(A\cap B\right)=1$ as only $08$ is the number with sum $8$ and product zero

$=n\left(B\right)=14$

$P\left(\frac{A}{B}\right)=\frac{n(A\cap B)}{n\left(B\right)}=\frac{1}{14}$