Question

The random variable X represents the number of heads that occur when you toss 4 coins at the same time what is the probability distribution

Answer 1

Four coins are tossed. Therefore sample space contain $2^4$ i.e $16$ points.

Let $S$ be the sample space.

Then $S=$ { $HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT$

$THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT$ }

Let $X$ be the random variable representing the number of heads that occur.

Then $X$ can take the values $0,1,2,3,4.$

Now corresponding to each points of random variable $Z$ we will get their probabilities.

$\therefore P(X=0)=$ ​probability of getting no head

$=\frac{1}{16}$

$P(X=1)=$ probability of getting one head

$=\frac{4}{16}=\frac{1}{4}$

$P(X=2)=$ probability of getting two head

$=\frac{6}{16}=\frac{3}{8}$

$P(X=3)=$ probability of getting three head

$=\frac{4}{16}=\frac{1}{4}$

$P(X=4)=$ probability of getting four head

$=\frac{1}{16}$

Let $F(x)$ be the distribution function of the random variable $X$ .

If $-\infty<x<0,$ $F(x)=0$

If $0\leq x<1,$ $F(x)=P(X=0)=\frac{1}{16}$ If $1\leq x<2, F(x)=[P(X=0)+P(X=1)]=(\frac{1}{16}+\frac{4}{16})=\frac{5}{16}$

If $2\leq x<3, F(x)=[P(X=0)+P(X=1)+P(X=2)]=(\frac{1}{16}+\frac{4}{16}+\frac{6}{16})=\frac{11}{16}$

If $3\leq x<4, F(x)=[P(X=0)+P(X=1)+P(X=2)+P(X=3)]=(\frac{1}{16}+\frac{4}{16}+\frac{6}{16}+\frac{4}{16})=\frac{15}{16}$

If $4\leq x<\infty , F(x)=[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)]$

$=(\frac{1}{16}+\frac{4}{16}+\frac{6}{16}+\frac{4}{16}+\frac{1}{16})=1$

Which is the required distribution function.