The probability distribution of a random variable X is as follows… - Vidyadip

Question

The probability distribution of a random variable $X$ is as follows:
$
p(x)=\left\{\begin{array}{ll}
K(x-1) ; & x=2,3 \\
K ; & x=4 \\
K(6-x) ; & x=5
\end{array}\right.
$
Find the value of constant $k$ and the probability of the event that variable $X$ assumes even numbers.

✓

Answer

The probability distribution of a random variation is given as follows:
$
\begin{aligned}
& p(x)=k(x-1) ; x=2,3 \\
& \therefore p(2)=k(2-1)=k \\
& p(3)=k(3-1)=2 k \\
& p(x)=k ; x=4 \\
& \therefore p(4)=k \\
& p(x)=k(6-x)=k \\
& \therefore p(5)=k(6-5)=k
\end{aligned}
$
By condition of probability distribution, we must have
$
\begin{aligned}
& \Sigma p(x)=1 \\
& \therefore p(2)+p(3)+p(4)+p(5)=1 \\
& \therefore k+2 k+k+k=1 \\
& \therefore 5 k=1 \\
& \therefore k=\frac{1}{5}
\end{aligned}
$
Probability that variable $X$ assumes even numbers :
$
\begin{aligned}
& P[X=2,4] \\
& =p(2)+p(4) \\
& =k+k=2 k \text { (Putting } k=\frac{1}{5} \text { ) } \\
& p(2)+p(4)=2 \times \frac{1}{5}=\frac{2}{5}
\end{aligned}
$
Hence, the probability that the variable $X$ assumes even number obtained is $\frac{2}{5}$.

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