Two coins are tossed simultaneously. Describe the sample space S. Find the probability of getting:

(i) two heads

(ii) at least one head

(iii) at most one head

(iv) exactly one head

(v) no head

When you toss two coins simultaneously, we get either both heads, first head second tail, first tail second head, both tails.

S = {HH, HT, TH, TT}

Total number of outcomes = 4

(i) Two heads

Number of favorable outcomes (two heads) = 1 (HH)

∴ P(getting two heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{4}$

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

(ii) At least one head

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

∴ P(getting at least one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

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

(iii) At most one head

Number of favorable outcomes (Getting at most one head) = 3 (HT, TH and TT)

∴ P(getting at most one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

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

(iv) Exactly one head

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

∴ P(getting exactly one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{4} = \dfrac{1}{2}.$

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

(v) No head

Number of favorable outcomes (Getting no head) = 1 (TT)

∴ P(getting no head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{4}.$

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