Three coins are tossed once. Find the probability of getting :

(i) 3 heads

(ii) exactly two heads

(iii) atleast 2 heads.

On tossing three coins.

Possible outcomes = {(HHH), (HHT), (HTT), (THH), (TTH), (TTT), (THT), (HTH)}.

No. of possible outcomes = 8.

(i) Favourable outcomes for getting 3 heads = (HHH).

∴ No. of favourable outcomes = 1.

P(getting three heads) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{8}$.

Hence, probability of getting three heads = $\dfrac{1}{8}$.

(ii) Favourable outcomes for getting exactly 2 heads = {(HHT), (THH), (HTH)}.

∴ No. of favourable outcomes = 3.

P(getting exactly two heads) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{8}$.

Hence, probability of getting exactly 2 heads = $\dfrac{3}{8}$.

(iii) Favourable outcomes for getting at least 2 heads = {(HHT), (THH), (HTH), (HHH)}.

∴ No. of favourable outcomes = 4.

P(getting atleast two heads) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{8} = \dfrac{1}{2}$.

Hence, probability of getting atleast 2 heads = $\dfrac{1}{2}$.