A random variable X has the following probability distribution: beginmatrixX :&1&2&3&4&5 P(X) :&K2&2K&K&2K&5K2 endmatrix Then P(X 2) is equal to :

Question

A random variable X has the following probability distribution: beginmatrixX :&1&2&3&4&5 P(X) :&K2&2K&K&2K&5K2 endmatrix Then P(X > 2) is equal to : (A) (7/12) (B) (23/36) (C) (1/36) (D) (1/6)

Tardigrade Admin · Accepted Answer

∑ P(X) = 1 ⇒ K2 + 2K + K + 2K + 5K2 = 1 ⇒ 6K2 + 5K - 1 = 0 ⇒ (6K - 1) (K + 1) = 0 ⇒ K = -1 (rejected) ⇒ K = (1/6) P(X > 2) = K + 2K + 5K2 =(23/36)

Q. A random variable X has the following probability distribution:
$X : P (X) : 1 K^{2} 2 2 K 3 K 4 2 K 5 5 K^{2}$
Then $P (X > 2)$ is equal to :

$\frac{7}{12}$

$\frac{23}{36}$

$\frac{1}{36}$

$\frac{1}{6}$

$\sum P (X) = 1 \Rightarrow K^{2} + 2 K + K + 2 K + 5 K^{2} = 1$
$\Rightarrow 6 K^{2} + 5 K - 1 = 0 \Rightarrow (6 K - 1) (K + 1) = 0$
$\Rightarrow K = - 1 (re j ec t e d) \Rightarrow K = \frac{1}{6}$
$P (X > 2) = K + 2 K + 5 K^{2} = \frac{23}{36}$

Q. A random variable X has the following probability distribution: X:P(X):​1K2​22K​3K​42K​55K2​ Then P(X>2) is equal to :

127​

3623​

361​

61​