Two coins are tossed together. Find the probability of getting :

(i) exactly one tail

(ii) at least one head

(iii) no head

(iv) at most one head

When two coins are tossed together, the total number of possible outcomes = 4 (i.e. HH, HT, TH and TT)

(i) Number of favourable outcomes (Getting exactly one tail) = 2 (HT and TH)

P(Getting exactly one tail) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{2}{4}$

= $\dfrac{1}{2}$

Hence, the probability of getting exactly one tail is $\dfrac{1}{2}$.

(ii) Number of favourable outcomes (Getting at least one head) = 3 (HH, HT and TH)

P(Getting at least one head) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{3}{4}$

Hence, the probability of getting at least one head is $\dfrac{3}{4}$.

(iii) Number of favourable outcomes (Getting no head) = 1 (TT)

P(Getting no head) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{1}{4}$

Hence, the probability of getting no head is $\dfrac{1}{4}$.

(iv) Number of favourable outcomes (Getting at most one head) = 3 (HH, HT and TT)

P(Getting at most one head) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$

= $\dfrac{3}{4}$

Hence, the probability of getting at most one head is $\dfrac{3}{4}$.