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

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Question

(d) Total number of ways $ = (HH,\,HT,\,TH,\,TT)$

$P$ (head on first toss) $ = \frac{2}{4} = \frac{1}{2} = P(A)$

$P$ (head on second toss) $ = \

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(d) Total number of ways $ = (HH,\,HT,\,TH,\,TT)$

$P$ (head on first toss) $ = \frac{2}{4} = \frac{1}{2} = P(A)$

$P$ (head on second toss) $ = \frac{2}{4} = \frac{1}{2} = P(B)$

$P$ (head on both toss) $ = \frac{1}{4} = P(A \cap B)$

Hence required probability is,

$P(A \cup B) = P(A) + P(B) - P(A \cap B) $

$= \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = \frac{3}{4}$ .

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

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