Consider the experiment of throwing a die. If | vedclass

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(A) The sample space $S$ of the experiment is defined by the outcomes of the first throw and the subsequent action.If the first throw is $1, 2, 4, 5$

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(A) The sample space $S$ of the experiment is defined by the outcomes of the first throw and the subsequent action.If the first throw is $1, 2, 4, 5$ (probability $4/6 = 2/3$),we toss a coin (outcomes $H, T$ with probability $1/2$ each).If the first throw is $3, 6$ (probability $2/6 = 1/3$),we throw the die again (outcomes $1, 2, 3, 4, 5, 6$ with probability $1/6$ each).The total sample space is:$S = \{(1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$.Let $A$ be the event that the coin shows a tail: $A = \{(1, T), (2, T), (4, T), (5, T)\}$.Let $B$ be the event that at least one die shows a $3$: $B = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)\}$.Since $A \cap B = \phi$,the intersection of events $A$ and $B$ is empty.Therefore,$P(A \cap B) = 0$.The conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{P(B)} = 0$.

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

Consider the experiment of throwing a die. If a multiple of $3$ comes up,throw the die again,and if any other number comes,toss a coin. Find the conditional probability of the event 'the coin shows a tail',given that 'at least one die shows a $3$'.

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