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A coin is tossed three times in succession. If $E$ is the event that there are at least two heads and $F$ is the event in which first throw is a head, then $P\left(\frac{E}{F}\right)=$
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The correct answer is:
$\frac{3}{4}$
$S=\{H H H, H H T, H T H, T H H, H T T, T H T, T T H, T T T$
$n(E)=4, n(F)=4$ and $n(E \cap F)=3$
$\therefore P\left(\frac{E}{F}\right)$ $=\frac{P(E \cap F)}{P(F)}=\frac{3 / 8}{4 / 8}=\frac{3}{4}$.
$n(E)=4, n(F)=4$ and $n(E \cap F)=3$
$\therefore P\left(\frac{E}{F}\right)$ $=\frac{P(E \cap F)}{P(F)}=\frac{3 / 8}{4 / 8}=\frac{3}{4}$.
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