One ticket is selected at random from (50) tickets numbered... | Mathematics | SolveeIt | Solveeit

Today, August 23

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

$\frac{1}{14}$

Explanation

Solution

$S=\left\\{00, 01, 02, \dots., 49\right\\}$ 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 $B=\left\\{00, 01, 02, 03, \dots., 09, 10, 20, 30, 40\right\\}$ $A nB =\left\\{8\right\\}$ Required probability $= P\left(A/B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{\frac{1}{50}}{\frac{14}{50}}=\frac{1}{14}$