A random variable x has the following probability distribution.

X	0	1	2	3	4	5	6
P(X)	k	3k	5k	7k	9k	11k	13k

Find the value of k and P(X>=2)

Hint: In this question we have to use the concept that the sum of each possible case in the probability distribution should be equal to 1. This means that-
$P(A_1) + P(A_2) +...+ P(A_n) = 1$

Complete step-by-step solution -
In the given probability distribution, there are 7 possible cases each with their own probability. The sum of all P(X) is equal to 1. So we can write that-
P(0) + P(1) + P(2) + P(3) +P(4) +P(5) + P(6) = 1
k + 3k + 5k + 7k + 9k + 11k + 13k = 1
49k = 1
$\mathrm k=\dfrac1{49}$
Now, we have to find the value of P(X>=2).
P(X>=2) = P(2) + P(3) +P(4) +P(5) + P(6)
= 5k + 7k + 9k + 11k + 13k
= 45k
Substituting the value of k we get-
$=\dfrac{45}{49}$
This is the required answer.

Note: In this problem, the most common mistake is that students do not include P(2) in P(X>=2). One should read the problem carefully as there is a difference between the greater than sign and the greater than or equal to sign.