One ticket is selected at random from 50 tick | vedclass

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(D) Let $S = \{00, 01, 02, \ldots, 49\}$ be the sample space. The total number of tickets is $50$.Let $A$ be the event that the sum of the digits on t

Vedclass · Accepted Answer

$\frac{1}{14}$

Vedclass · Answer

(D) Let $S = \{00, 01, 02, \ldots, 49\}$ be the sample space. The total number of tickets is $50$.Let $A$ be the event that the sum of the digits on the selected ticket is $8$.$A = \{08, 17, 26, 35, 44\}$.Let $B$ be the event that the product of the digits is zero.This occurs if at least one digit is $0$. The tickets are $\{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\}$.There are $14$ such tickets,so $n(B) = 14$.We need to find the conditional probability $P(A|B) = \frac{n(A \cap B)}{n(B)}$.$A \cap B$ is the set of tickets where the sum of digits is $8$ $AND$ the product of digits is $0$.Looking at set $A = \{08, 17, 26, 35, 44\}$,only $08$ has a product of digits equal to $0$ $(0 	imes 8 = 0)$.Thus,$A \cap B = \{08\}$,so $n(A \cap B) = 1$.Therefore,$P(A|B) = \frac{1}{14}$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

One ticket is selected at random from $50$ tickets numbered $00, 01, 02, \ldots, 49$. The probability that the sum of the digits on the selected ticket is $8$,given that the product of these digits is zero,equals:

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