Question

A fair coin is tossed four times. Let \[X\] denote a string of heads occurring, find the probability distribution, mean and variance of $X$ .

Answer 1

Hint:Probability distribution: A probability distribution is a statistical function that describes all the possible value and likelihoods that a random variable can take within a given range.
Mean: Mean is the simple mathematical average of a set of two or more numbers.
mean \[ = \sum {\,{P_i}{X_i}} \]
Variance: Variance is the expected value of the squared variation of a random variable from its mean value, in probability and statistic.
 $ = \sum {\,{P_i}{X_i}^2 - {{(Mean)}^2}} $

Complete step by step answer:
A fair coin is tossed four times. Let $X$ denote the number of heads occurring.
As we know, If a coin is tossed \[4\] times, then the possible outcomes are: $HHHH,HHHT,HHTT,HTTT,THHH,\,....$
Total number of outcomes of this will be $ = 16$ .
 $X$ can take the values $0,1,2,3$ and $4$ .
Let when there is n number of head in $4$ coin;
Then, number of such combination will be $^4{C_n} = \dfrac{{4!}}{{n!(4 - n)!}}$
Here, $n > 4$
Now, we can write;
When zero head;
$P\left( {X = 0} \right) = P\left( {0\,head} \right) = \dfrac{1}{{16}}$
When one head;
 $P\left( {X = 1} \right) = P\left( {1\,head} \right) = \dfrac{4}{{16}}$
When two head;
 $P\left( {X = 2} \right) = P\left( {2\,heads} \right) = \dfrac{6}{{16}}$
When three head;
 $P\left( {X = 3} \right) = P\left( {3\,heads} \right) = \dfrac{4}{{16}}$
When four head;
$P\left( {X = 4} \right) = P\left( {4\,heads} \right) = \dfrac{1}{{16}}$
Thus, the probability distribution of $X$ is given by

$X$	$P(X)$
$0$	$\dfrac{1}{{16}}$
$1$	$\dfrac{4}{{16}}$
$2$	$\dfrac{6}{{16}}$
$3$	$\dfrac{4}{{16}}$
$4$	$\dfrac{1}{{16}}$

Computation of mean and variance. We will create table as per given below;

${X_i}$	${P_i}$	${P_i}{X_i}$	${P_i}{X_i}^2$
$0$	$\dfrac{1}{{16}}$	$0$	$0$
$1$	$\dfrac{4}{{16}}$	$\dfrac{4}{{16}}$	$\dfrac{4}{{16}}$
$2$	$\dfrac{6}{{16}}$	$\dfrac{{12}}{{16}}$	$\dfrac{{24}}{{16}}$
$3$	$\dfrac{4}{{16}}$	$\dfrac{{12}}{{16}}$	$\dfrac{{36}}{{16}}$
$4$	$\dfrac{1}{{16}}$	$\dfrac{4}{{16}}$	$1$
		$\sum {{P_i}{X_i} = 2} $	$\sum {{P_i}{X_i}^2 = 5} $

Mean \[ = \sum {\,{P_i}{X_i}} = 2\]
Variance $ = \sum {\,{P_i}{X_i}^2 - {{(Mean)}^2}} $
Keeping value in it. We get,
 $ = 5 - 4$
 $ = 1$

Note: Probability is applied in everyday life in risk assessment and modeling. The insurance industry and markets use actuarial science to determine pricing and make trading decisions. Government apply probabilistic methods in environmental regulation, entitlement analysis and financial regulation.