Question

A random variable X has the following probability distribution

 then P (X ≥ 2) =?

• A. \(\frac {1}{49}\)
• B. \(\frac {45}{49}\)
• C. \(\frac {40}{49}\)
• D. \(\frac {15}{49}\)

Answer 1

The Correct Option is B

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:
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
k = \(\frac {1}{49}\)
Now we have to calculate the probability P(X ≥ 2):
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
Here,
P(X = 2) = 5k = 5\((\frac {1}{49})\) = \(\frac {5}{49}\)
P(X = 3) = 7k = 7\((\frac {1}{49})\) = \(\frac {7}{49}\)
P(X = 4) = 9k = 9\((\frac {1}{49})\) = \(\frac {9}{49}\)
P(X = 5) = 11k = 11\((\frac {1}{49})\) = \(\frac {11}{49}\)
P(X = 6) = 13k = 13\((\frac {1}{49})\) = \(\frac {13}{49}\)
Adding these probabilities:
P(X ≥ 2) = \(\frac {5}{49}\) + \(\frac {7}{49}\) + \(\frac {9}{49}\) + \(\frac {11}{49}\) + \(\frac {13}{49}\) = \(\frac {45}{49}\) 
Therefore, the correct option is (B) \(\frac {45}{49}\)