One ticket is selected at random from 50 tickets numbered $\{00,01,02, \ldots . ., 49\}$. Then the…

One ticket is selected at random from 50 tickets numbered $\{00,01,02, \ldots . ., 49\}$. Then the probability that the sum of the igits on the selected ticket is 8 , given that the product of these digits is zero, is
  1. $\frac{1}{50}$
  2. $\frac{1}{14}$
  3. $\frac{14}{50}$
  4. $\frac{1}{10}$

Solution

$\begin{aligned} & S=\{00,01,02, \ldots \ldots ., 49\}, n(S)=50 \\ & E_1=\{08,17,26,35,44\}, n\left(E_1\right)=5 \\ & E_2=\{00,01,02,03,04,05,06,07,08,09,10,20,30,40\}, n\left(E_2\right)=14 \\ & E_1 \cap E_2=\{08\}, n\left(E_1 \cap E_2\right)=1 \\ & \text { Required probability }=p\left(E_1 / E_2\right)=\frac{n\left(E_1 \cap E_2\right)}{n\left(E_2\right)}=\frac{1}{14}\end{aligned}$

Asked in: MHT CET 2022 (10 Aug Shift 1)

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