This is a multi-part question in the textbook answered separately on Clay6.com

- Once we write down the sample space for the problem we can find the number of heads or tails for the favorable outcomes in each case and find their probabilities and arrange as a probability distribution.

Given: one coin is tossed twice, we can write the sample space as follows: S = {HH, HT, TH, TT}.

Since all these 4 events are equally likely, P(HH) = P(HT) = P(TH) = P(TT) = $\large\frac{1}{4}$

Let X be the number of heads in S such that $\rightarrow$ X(HH) = 2, X(HT) = 1, X(TH) = 1 and X(TT) = 0. It follows that:

P(X = 0) = P (TT) = $\large\frac{1}{4}$

P(X = 1) = P(TH, HT) = $\large\frac{2}{4} = \frac{1}{2}$

P(X = 2) = P(HH) = $\large\frac{1}{4}$

We can then construct the probability distribution table as follows:

$\begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{1}{4} &\large \frac{1}{2} &\large \frac{1}{4} \end{matrix}$

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