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

When a coins tossed twice, the all possible results are HH, HT, TH, TT.
Total number of possible out comes $=4$
Probability of an event $=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$
(i) Exactly one head,
Number of possible out comes =2 ( HT, TH).
Probability of an event getting exactly one head$=\frac{2}{4}=\frac{1}{2}$
(ii) Exactly one tail,

Number of possible out comes =2 ( HT, TH).

Probability of an event getting exactly one tail$=\frac{2}{4}=\frac{1}{2}$
(iii) Two tails,

Number of possible out comes =1 ( TT).

Probability of an event getting exactly one head$=\frac{1}{4}$
(iii) Two heads,

Number of possible out comes =1 (HH).

Probability of an event getting exactly one head$=\frac{1}{4}$

∴ \therefore=24=12 =\frac{2}{4} = \frac{1}{2