Three fair coins are tossed and either heads(H) or tails(T) are observed for each coin.

1. List the sample points for the experiment.
2. Assign probabilities to the sample points.
3. Determine the probability of observing each of the following events:

A= {Three heads are observed}

B= {Exactly two heads are observed}

C= {At least two heads are observed}

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$\frac{\mathbf{1}}{\mathbf{2}}$.