Out of 50 tickets numbered 00, 01, 02, dots, | vedclass

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Question

(C) Let $S$ be the sample space of $50$ tickets: $S = \{00, 01, \dots, 49\}$.Event $A$ is the set of tickets where the sum of digits is $8$: $A = \{08

Vedclass · Accepted Answer

$1/14$

Vedclass · Answer

(C) Let $S$ be the sample space of $50$ tickets: $S = \{00, 01, \dots, 49\}$.Event $A$ is the set of tickets where the sum of digits is $8$: $A = \{08, 17, 26, 35, 44\}$.Event $B$ is the set of tickets where the product of digits is $0$. This occurs if at least one digit is $0$: $B = \{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\}$.The number of elements in $B$ is $n(B) = 14$.The intersection $A \cap B$ is the set of tickets where the sum of digits is $8$ $AND$ the product of digits is $0$. Looking at set $A$,only $08$ satisfies this condition: $A \cap B = \{08\}$.Thus,$n(A \cap B) = 1$.The conditional probability $P(A|B)$ is given by:$P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{1}{14}$.

List-$I$	List-$II$
$A$. $P(\frac{A}{B})$	$(i)$. $\frac{2}{15}$
$B$. $P(\bar{B})$	$(ii)$. $\frac{4}{15}$
$C$. $P(A \cap \bar{B})$	$(iii)$. $\frac{8}{15}$
$D$. $P(B \cap \bar{A})$	$(iv)$. $\frac{2}{3}$
	$(v)$. $\frac{3}{7}$

Out of $50$ tickets numbered $00, 01, 02, \dots, 49$,one ticket is drawn at random. Let $A$ be the event that the sum of the digits on the ticket is $8$,and $B$ be the event that the product of the digits is $0$. Find the conditional probability $P(A|B)$.

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