Consider the experiment of tossing a coin. If | vedclass

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(A) The outcomes of the experiment can be represented using a tree diagram.The sample space $S$ of the experiment is:$S = \{(H, H), (H, T), (T, 1), (T

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$\frac{2}{9}$

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(A) The outcomes of the experiment can be represented using a tree diagram.The sample space $S$ of the experiment is:$S = \{(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\}$where $(H, H)$ denotes both tosses result in heads,and $(T, i)$ denotes the first toss results in a tail and the number $i$ appears on the die for $i \in \{1, 2, 3, 4, 5, 6\}$.The probabilities assigned to these elementary events are:$P(H, H) = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$$P(H, T) = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$$P(T, i) = \frac{1}{2} 	imes \frac{1}{6} = \frac{1}{12}$ for each $i \in \{1, 2, 3, 4, 5, 6\}$.Let $F$ be the event that 'there is at least one tail' and $E$ be the event 'the die shows a number greater than $4$'.$F = \{(H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\}$$E = \{(T, 5), (T, 6)\}$$E \cap F = \{(T, 5), (T, 6)\}$$P(F) = P(H, T) + P(T, 1) + P(T, 2) + P(T, 3) + P(T, 4) + P(T, 5) + P(T, 6)$$P(F) = \frac{1}{4} + 6 	imes \frac{1}{12} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$$P(E \cap F) = P(T, 5) + P(T, 6) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/6}{3/4} = \frac{1}{6} 	imes \frac{4}{3} = \frac{4}{18} = \frac{2}{9}$

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

Consider the experiment of tossing a coin. If the coin shows head,toss it again but if it shows tail,then throw a die. Find the conditional probability of the event that 'the die shows a number greater than $4$' given that 'there is at least one tail'.

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