Question

Find the mean number of heads in three tosses of fair coin.

Answer 1

Let X denote the success of getting heads.

Therefore, the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that X can take the value of 0, 1, 2, or 3.

P(X=0)=P(TTT)

=P(T).P(T).P(T)

=$\frac{1}{2}X\frac{1}{2}X\frac{1}{2}$=$\frac{1}{8}$

∴ P (X = 1) = P (HHT) + P (HTH) + P (THH)

$\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}$=$\frac{3}{8}$

∴P(X = 2) = P (HHT) + P (HTH) + P (THH)

$\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}$=$\frac{3}{8}$

∴P(X=3)=P(HHH)=$\frac{1}{2}X\frac{1}{2}X\frac{1}{2}$

=$\frac{1}{8}$

Therefore, the required probability distribution is as follows.

X	0	1	2	3
P(X)	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Mean of XX(X),μ=∑XiP(Xi)

=$0×\frac{1}{8}+1×\frac{3}{8}+2×\frac{3}{8}+3×\frac{1}{8}$

=$\frac{3}{8}+\frac{3}{8}+\frac{3}{8}$

=$\frac{3}{2}=1.5$