A coin is tossed twice. Find the probability of getting :

(i) exactly one head

(ii) exactly one tail

(iii) two tails

(iv) two heads

When a coin is tossed, each toss has 2 possible outcomes : head (H) or tail (T).

The number of possible outcomes when a coin is tossed n times = 2ⁿ

For this case, n = 2

So, 2² = 4

The number of possible outcomes when a coin is tossed 2 times is 4.

(i) Number of favourable outcomes (Getting exactly one head) = 2 (HT, TH)

Total number of outcomes = 4

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

= $\dfrac{2}{4}$

= $\dfrac{1}{2}$

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

(ii) Number of favourable outcomes (Getting exactly one tail) = 2 (HT, TH)

Total number of outcomes = 4

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

= $\dfrac{2}{4}$

= $\dfrac{1}{2}$

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

(iii) Number of favourable outcomes (Getting two tails) = 1 (TT)

Total number of outcomes = 4

P(Getting two tails) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{1}{4}$

Hence, the probability of getting two tails is $\dfrac{1}{4}$.

(iv) Number of favourable outcomes (Getting two heads) = 1 (HH)

Total number of outcomes = 4

P(Getting two heads) = $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

= $\dfrac{1}{4}$

Hence, the probability of getting two heads is $\dfrac{1}{4}$.