Question

Find the probability distribution of

number of heads in four tosses of a coin.

Answer 1

When a coin is tossed four times, the sample space is

$\left\{\begin{array}{cccccccc}HHHH& HHHT& HHTH& HTHT& HTTH& HTTT& THHH& HTHH\\ THHT& THTH& HHTT& TTHH& TTHT& TTTH& THTT& TTTT\end{array}\right\}$

which contains 16equally likely sample points.

Let X be the random variable, which represent the number of heads. It can be seen that X can take the value of 1,2,3, or 4.

P(X=0)=P (no head shows up) = P{TTTT} =$\frac{1}{16}$,

P(X=1)= P (one head and three tails show up)

= P(HTTT,THTT,TTHT,TTTH)=$\frac{4}{16}=\frac{1}{4},$

P(X=2) =P (two heads and two tails show up)

= P({HHTT,HTHT,HTTH,THHT,THTH,TTHH}) =$\frac{6}{16}=\frac{3}{8}$,

P(X=3) =P (three heads and one tail show up)

= P({HHHT,HHTH,HTHH,THHH})= $\frac{4}{16}=\frac{1}{4}$

and P(X=4)=P (four heads show up) = P({HHHH}) =$\frac{1}{16}$

Thus, the probability distribution is as follows:

$\begin{array}{cccccc}X& 0& 1& 2& 3& 4\\ P\left(X\right)& \frac{1}{16}& \frac{1}{4}& \frac{3}{8}& \frac{1}{4}& \frac{1}{16}\end{array}$