Question

A random variable $X$ has the following probability distribution:
$\begin{array}{cc}X& P\left(X\right)\\ 0& 0\\ 1& k\\ 2& 2k\\ 3& 2k\\ 4& 3k\\ 5& k^2\\ 6& 2k^2\\ 7& 7k^2+k\end{array}$

Determine:
$\left(i\right)k$
$\left(ii\right)P(X<3)$
$\left(iii\right)P(X>6)$
$\left(iii\right)P(0<X<3)$

Answer 1

Given probability distribution:
$\begin{array}{ccccccccc}X& 0& 1& 2& 3& 4& 5& 6& 7\\ P\left(x\right)& 0& k& 2k& 2k& 3k& k^2& 2k^2& 7k^2+k\end{array}$
$\text{Part}\left(i\right)$
We know that,
Sum of probabilites in a probability
distribution=$1$

$\Rightarrow {\sum }_0^{7}P\left(X\right)=1$

$\Rightarrow P(X=0)+P(X=1)+P(X=2)$
$+P(X=3)+P(X=4)+P(X=5)$
$+P(X=6)+P(X=7)=1$
$\Rightarrow 0+k+2k+2k+3k+k^2+2k^2$
$+7k^2+k=1$
$\Rightarrow 9k+10k^2=1$
$\Rightarrow 10k^2+9k-1=0$
$\Rightarrow 10k^2+10k-k-1=0$
$\Rightarrow 10k(k+1)-1(k+1)=0$
$\Rightarrow (10k-1)(k+1)=0$
So, $k=\frac{1}{10}\&k=-1$
Since $k$ is a probability, it cannot be negative,

Hence,$k=\frac{1}{10}$

$Part\left(ii\right)$
From given table,
$P(X<3)=P(X=0)+P(X=1)+P(X=2)$
$P(X<3)=0+k+2k$
$P(X<3)=3k$

$P(X<3)=3\times \frac{1}{10}$ $(\because k=\frac{1}{10})$

$\therefore P(X<3)=\frac{3}{10}$

$Part\left(iii\right)$
$P(X>6)=P(X=7)$
$P(X>6)=7k^2+k$

$P(X>6)=7{\left(\frac{1}{10}\right)}^2+\frac{1}{10}$ $(\because k=\frac{1}{10})$

$P(X>6)=7\times \frac{1}{100}+\frac{1}{10}$

$P(X>6)=\frac{7+10}{100}$

$\therefore P(X>6)=\frac{17}{100}$

$Part\left(iv\right)$
$P(0<X<3)=P(X=1)+P(X=2)$

$\Rightarrow P(0<X<3)=k+2k$
$\Rightarrow P(0<X<3)=3k$

$\Rightarrow P(0<X<3)=3\times \frac{1}{10}$ $(\because k=\frac{1}{10})$

$\therefore P(0<X<3)=\frac{3}{10}$