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

Step 1: Find the sample space:

Sample space for tossing three coins will be,

$$\begin{gathered} n\left(S\right)={\text{(number of possible outcomes of tossing a coin)}}^{\text{number of coins tossed}} \\ n\left(S\right)={\left(2\right)}^3 \\ n\left(S\right)=8 \end{gathered}$$

The sample space are $\{HHH,TTT,HHT,HTH,THH,TTH,THT,HTT\}$

Step 2: Find the probability of getting exactly one head:

Thus, the favorable outcomes will be,

$\{TTH,THT,HTT\}$

Therefore, the number of favorable outcomes will be $3$

The formula for probability will be,

$\text{Probability =}\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Thus, the probability will be,

$\text{Probability=}\frac{3}{8}$

Step 3: Find the probability of getting at least one head and one tail:

Thus, the favorable outcomes will be,

$\{HHT,HTH,THH,TTH,THT,HTT\}$

Therefore, the number of favorable outcomes will be $6$

The formula for probability will be,

$\text{Probability =}\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Thus, the probability will be,

$$\begin{gathered} \text{Probability=}\frac{6}{8} \\ \text{=}\frac{3}{4} \end{gathered}$$

Hence, the probability of :

(i) getting exactly one head $=\frac{3}{8}$

(ii) getting at least one head and one tail$=\frac{3}{4}$