Question 354 of 417MediumMCQ
A random variable X has the following
probability distribution :
X:
1
2
3
4
5
P(X):
K2
2K
K
2K
5K2
Then P(X > 2) is equal to :
Solutions · 1
Official solution
\[\sum\limits_{i = 1}^5 {P(X)} \] = 1
\[ \Rightarrow \] K2
+ 2K + K + 2K + 5K2
= 1
\[ \Rightarrow \] 6K2
+ 5K -- 1 = 0
\[ \Rightarrow \] (6K - 1)(k + 1) = 0
\[ \Rightarrow \] K = \[{1 \over 6}\] and K = -1(rejected)
\[ \therefore \] P(X > 2)
= K + 2K + 5K2
= \[{1 \over 6} + {2 \over 6} + {5 \over {36}}\]
= \[{{23} \over {36}}\] Answer the question to unlock