# Find the probability distribution of number of heads in four tosses of a coin.

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

Toolbox:
• 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: 4 tosses of a coin, the sample space can be represented as follows: S = $\begin{Bmatrix} HHHH & HHHT & HHTH &HHTT \\ HTHH &HTHT &HTTH &HTTT \\ THHH & THHT &THTH &THTT \\ TTHH &TTHT &TTTH & TTTT \end{Bmatrix}$
All these events are equally likely and have the same probability = $\large\frac{1}{16}$
Let X be the number of heads in those tosses. We can see that X can take the values of 0,1,2,3 or 4. It follows that:
P (X = 0) = P (TTTT) = $\large\frac{1}{16}$
P (X = 1) = P (HTTT, THTT, TTHT, TTTH) = $\large\frac{4}{16} = \frac{1}{4}$
P (X = 2) = P (HHTT, HTHT, HTTH, TTHH, THHT, THTH) = $\large\frac{6}{16} = \frac{3}{8}$
P (X = 3) = P (HHHT, HHTH, HTHH, THHH) = $\large\frac{4}{16} = \frac{1}{4}$
P (X = 4) = P (HHHH) = $\large\frac{1}{16}$
Therefore we can write the probability distribution as follows:
$\begin{matrix} \textbf{X} & 0 &1 &2&3&4 \\ \textbf{P(X)} &\large \frac{1}{16} &\large \frac{1}{4} &\large \frac{3}{8} &\large\frac{1}{4}&\large\frac{1}{16}\end{matrix}$