Question

A coin is tossed twice. Find the probability of getting:
(i) exactly one head
(ii) exactly one tail
(iii) two tails
(iv) two heads​

Answer 1

A coin is tossed twice.

This is an experiment

We will get

HH, HT, TT, TH

now,

total possible outcomes = 4

(i) exact one head

favourable outcome = HT, TH

number of favourable outcomes = 2

$Probability(exact \: one \: head) = \frac{no.of \: favourable \: outoes}{no. \: of \: possile \: out \: comes} \\ \\ Probability(exact \: one \: head) = \frac{2}{4} = \frac{1}{2}$

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(ii) exact one tail

favourable outcome = HT, TH

number of favourable outcomes = 2

$Probability(exact \: one \: tail) = \frac{no.of \: favourable \: outoes}{no. \: of \: possile \: out \: comes} \\ \\ Probability(exact \: one \: ail) = \frac{2}{4} = \frac{1}{2}$

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(iii) two tails

favourable outcome = TT

number of favourable outcomes = 1

$Probability(two \: tails) = \frac{no.of \: favourable \: outoes}{no. \: of \: possile \: out \: comes} \\ \\ Probability(two \: tails) = \frac{1}{4}$

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(iv) two heads

favourable outcome = HH

number of favourable outcomes = 1

$Probability(two \: heads) = \frac{no.of \: favourable \: outoes}{no. \: of \: possile \: out \: comes} \\ \\ Probability(two \: heads) = \frac{1}{4}$