Find P(E | F) when two coins are tossed once, | vedclass

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(A) When two coins are tossed once,the sample space $S$ is given by:$S = \{HH, HT, TH, TT\}$The event $E$ is that no tail appears,so $E = \{HH\}$.The

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(A) When two coins are tossed once,the sample space $S$ is given by:$S = \{HH, HT, TH, TT\}$The event $E$ is that no tail appears,so $E = \{HH\}$.The event $F$ is that no head appears,so $F = \{TT\}$.Therefore,the intersection of $E$ and $F$ is $E \cap F = \phi$,which means the number of elements in $E \cap F$ is $0$.The probability of event $F$ is $P(F) = \frac{n(F)}{n(S)} = \frac{1}{4}$.The conditional probability $P(E | F)$ is defined as:$P(E | F) = \frac{P(E \cap F)}{P(F)}$Since $P(E \cap F) = 0$,we have:$P(E | F) = \frac{0}{1/4} = 0$.

Find $P(E | F)$ when two coins are tossed once,where $E$ is the event that no tail appears and $F$ is the event that no head appears.

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