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14.Probability
easy
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 = $
A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$\frac{1}{8}$
D
$\frac{3}{4}$
Solution
(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) $
$= \frac{1}{2} + \frac{1}{2} – \frac{1}{4} = \frac{3}{4}$ .
Standard 11
Mathematics
