Question

A coin is tossed three times then find the probability of
(i) getting head on middle coin.
(ii) getting exactly one tail.
(iii) getting no tail.

Answer 1

Given:
A coin is tossed three times.
∴ Sample space = {HHH, THH, HTH, HHT, THT, TTH, HTT, TTT}
Or,
n(S) = 8

(i) Let X be the event of getting head on the middle coin.
X = {HHH, THH, HHT, THT}
∴ n(X) = 4
Thus, we have:
P(X) = $\frac{n\left(\mathrm{X}\right)}{n\left(\mathrm{S}\right)}$
$\Rightarrow$P(X) =$\frac{4}{8}=\frac{1}{2}$

(ii) Let Y be the event of getting exactly one tail.
Y = {THH, HTH, HHT}
∴ n(Y) = 3
Thus, we have:
P(Y) = $\frac{n\left(\mathrm{Y}\right)}{n\left(\mathrm{S}\right)}$
$\Rightarrow$P(Y) = $\frac{3}{8}$

(iii) Let Z be the event of getting no tail.
Z = {HHH}
∴ n(Z) = 1
Thus, we have:
P(Z) = $\frac{n\left(\mathrm{Z}\right)}{n\left(\mathrm{S}\right)}$
$\Rightarrow$P(Z) = $\frac{1}{8}$