91Ó°ÊÓ

While tossing a fair coin 3 times, the number of total outcomes will be equal to 2³ = 8. Let us denote H for observing a head and T for observing a tail,

Therefore, the Sample points are$\mathbf{=}\left\{\left(\mathbf{H}\mathbf{,}\mathbf{H}\mathbf{,}\mathbf{H}\right)\mathbf{,}\left(\mathbf{H}\mathbf{,}\mathbf{T}\mathbf{,}\mathbf{H}\right)\mathbf{,}\left(\mathbf{H}\mathbf{,}\mathbf{H}\mathbf{,}\mathbf{T}\right)\mathbf{,}\left(\mathbf{T}\mathbf{,}\mathbf{H}\mathbf{,}\mathbf{H}\right)\mathbf{,}\left(\mathbf{T}\mathbf{,}\mathbf{T}\mathbf{,}\mathbf{H}\right)\mathbf{,}\left(\mathbf{H}\mathbf{,}\mathbf{T}\mathbf{,}\mathbf{T}\right)\mathbf{,}\left(\mathbf{H}\mathbf{,}\mathbf{T}\mathbf{,}\mathbf{H}\right)\mathbf{,}\left(\mathbf{T}\mathbf{,}\mathbf{T}\mathbf{,}\mathbf{T}\right)\right\}$

$\mathbf{A}\mathbf{=}$Three heads are observed

One observes three heads on the coin only once, n(A) = 1, n(S) = total number of outcomes = 8

localid="1662212971078" $$\begin{gathered} \mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Favourable}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}} \\ \mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}} \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{8}} \end{gathered}$$

Therefore, the probability of getting 3 heads is $\frac{\mathbf{1}}{\mathbf{8}}$ .

$\mathbf{B}\mathbf{=}$Exactly two heads are observed

One observes exactly two heads (H, H, T), (H, T, H), and (T, H, H), thus n (B) = 3, n(S) = total number of outcomes = 8

localid="1662213016079" $$\begin{gathered} \mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Favourable}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}} \\ \mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}} \\ \mathbf{=}\frac{\mathbf{3}}{\mathbf{8}} \end{gathered}$$

Therefore, the probability of getting exactly 2 heads is $\frac{\mathbf{3}}{\mathbf{8}}$.

$\mathbf{C}\mathbf{=}$ At least two heads are observed.

One observes at least 2 heads (H,H,T), (H,T,H), and (T,H,H), and (H,H,H), thus n(A) = 4, n(S) = total number of outcomes = 8.

localid="1662213070863" $$\begin{gathered} \mathbf{P}\mathbf{\left(}\mathbf{C}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Favourable}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}} \\ \mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{C}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}} \\ \mathbf{=}\frac{\mathbf{4}}{\mathbf{8}} \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

Therefore, the probability of getting at least 2 heads is$\frac{\mathbf{1}}{\mathbf{2}}$.