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. $\frac{1}{14}$
• B. $\frac{1}{7}$
• C. $\frac{5}{14}$
• D. $\frac{1}{50}$

Answer 1

Answer: (A)

$S=\left\{00, 01, 02, \dots., 49\right\}$
Let A be the even that sum of the digits on the selected ticket is 8 then
$A=\left\{08, 17, 26, 35, 44\right\}$
Let B be the event that the product of the digits is zero
$B=\left\{00, 01, 02, 03, \dots., 09, 10, 20, 30, 40\right\}$
$A ∩B =\left\{8\right\}$
Required probability $= P\left(A/B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{\frac{1}{50}}{\frac{14}{50}}=\frac{1}{14}$