A coin is tossed twice. If events A and B are | vedclass

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(D) The sample space for tossing a coin twice is $S = \{(HH, HT, TH, TT)\}$.Event $A$ (head on first toss) $= \{(HH, HT)\}$,so $P(A) = \frac{2}{4} = \

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$\frac{3}{4}$

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(D) The sample space for tossing a coin twice is $S = \{(HH, HT, TH, TT)\}$.Event $A$ (head on first toss) $= \{(HH, HT)\}$,so $P(A) = \frac{2}{4} = \frac{1}{2}$.Event $B$ (head on second toss) $= \{(HH, TH)\}$,so $P(B) = \frac{2}{4} = \frac{1}{2}$.Event $A \cap B$ (head on both tosses) $= \{(HH)\}$,so $P(A \cap B) = \frac{1}{4}$.Using the addition theorem of probability:$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$P(A \cup B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4}$.

$A$ coin is tossed twice. If events $A$ and $B$ are defined as: $A = \text{head on first toss}$,$B = \text{head on second toss}$. Then the probability of $A \cup B = $

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