One ticket is selected at random from 50 tickets numbered $00,01,02, \ldots, 49$. Then the probability that…
One ticket is selected at random from 50 tickets numbered $00,01,02, \ldots, 49$. 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
$\frac{1}{14}$
$\frac{1}{7}$
$\frac{5}{14}$
$\frac{1}{50}$
Solution
$
S=\{00,01,02, \ldots .49\}
$
Let $A$ be the even that sum of the digits on the selected ticket is 8 then
$
A=\{08,17,26,35,44\}
$
Let $B$ be the event that the product of the digits is zero
$
B=\{00,01,02,03, \ldots .09,10,20,30,40\}
$
$A \cap B=\{8\}$
$
\text { Required probability }=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{50}}{\frac{14}{50}}=\frac{1}{14}
$