One ticket is selected at random from 50 tick | vedclass

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(B) Let $S$ be the sample space of $50$ tickets: $S = \{00, 01, 02, \ldots, 49\}$,so $n(S) = 50$.Let $E_1$ be the event that the sum of the digits is

Vedclass · Accepted Answer

$\frac{1}{14}$

Vedclass · Answer

(B) Let $S$ be the sample space of $50$ tickets: $S = \{00, 01, 02, \ldots, 49\}$,so $n(S) = 50$.Let $E_1$ be the event that the sum of the digits is $8$: $E_1 = \{08, 17, 26, 35, 44\}$,so $n(E_1) = 5$.Let $E_2$ be the event that the product of the digits is zero. This occurs if at least one digit is $0$: $E_2 = \{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\}$,so $n(E_2) = 14$.The intersection $E_1 \cap E_2$ is the set of tickets where the sum is $8$ and the product is $0$: $E_1 \cap E_2 = \{08\}$,so $n(E_1 \cap E_2) = 1$.The conditional probability is $P(E_1 | E_2) = \frac{n(E_1 \cap E_2)}{n(E_2)} = \frac{1}{14}$.

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,is

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