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(A) Let $H$ be the event of getting a head and $D$ be the event of getting a number greater than $4$ on a die.$P(H) = \frac{1}{2}$.The numbers greater

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$\frac{2}{3}$

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(A) Let $H$ be the event of getting a head and $D$ be the event of getting a number greater than $4$ on a die.$P(H) = \frac{1}{2}$.The numbers greater than $4$ on a die are ${5, 6}$,so $P(D) = \frac{2}{6} = \frac{1}{3}$.Since the events are independent,$P(H \cap D) = P(H) 	imes P(D) = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6}$.We need to find the probability of $H \cup D$,which is given by $P(H \cup D) = P(H) + P(D) - P(H \cap D)$.$P(H \cup D) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3+2-1}{6} = \frac{4}{6} = \frac{2}{3}$.

$A$ coin is tossed and a die is thrown. The probability that the outcome will be a head or a number greater than $4$ or both,is

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