Question

Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail.

Answer 1

When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
i.e. n (S) = 8

(i) Let E₁ = event of getting exactly two heads
Then E₁ = {HHT, HTH, THH}
i.e. n(E₁) = 3
$\therefore P\left(E_1\right)=\frac{n\left(E_1\right)}{n\left(S\right)}=\frac{3}{8}$

(ii) Let E₂ = event of getting at least two heads
Then E₂ = {HHH, HHT, HTH, THH}
i.e. n (E₂) = 4
$\therefore P\left(E_2\right)=\frac{n\left(E_2\right)}{n\left(S\right)}=\frac{4}{8}=\frac{1}{2}$

(iii) Let E₃ = event of getting at least one head and one tail
Then E₃ = {HHT, HTH, HTT, THH, THT, TTH}
i.e. n(E₃) = 6
$\therefore P\left(E_3\right)=\frac{n\left(E_3\right)}{n\left(S\right)}=\frac{6}{8}=\frac{3}{4}$