Question:
Two coins are tossed simultaneously. Find the probability of getting
(i) exactly 1 head
(ii) at most 1 head
(iii) at least 1 head
Solution:
Number of favourable outcomes = 2
When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH and TT.
Total number of possible outcomes = 4
(i) Let E be the event of getting exactly one head.
Then, the favourable outcomes are HT and TH.
Number of favourable outcomes = 2
$\therefore P($ getting exactly 1 head $)=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}=\frac{2}{4}=\frac{1}{2}$
(ii) Let E be the event of getting at most one head.
Then, the favourable outcomes are HT, TH and TT.
Number of favourable outcomes = 3
$\therefore P$ (getting at most 1 head) $=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}=\frac{3}{4}$
(iii) Let E be the event of getting at least one head.
Then, the favourable outcomes are HT, TH and HH
Number of favourable outcomes $=3$
$\therefore P($ getting at least 1 head) $)=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}=\frac{3}{4}$
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