Question

The following is probabilty distribution of r.v X.

x	1	2	3	4	5	6
p(x)	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$	$\frac{k}{6}$

then value of k is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

We know that, Suppose a random variable $X$ may take $k$ different values, with the probability that $X=x_i$ defined to be $P(X=x_i)=p_i$. The probabilities $p_i$ must satify the following:

1. $0\le p_i\le 1$ for each $i$.
   
2. $p_1+p_2+...+p_k=1$ that is $\sum p_i=1$

Thus we have $\sum p_i=1$

$\Rightarrow p_1+p_2+p_3+p_4+p_5+p_6=1$

$\Rightarrow \frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}+\frac{k}{6}=1$

$\Rightarrow \frac{6k}{6}=1$

$\Rightarrow k=1$