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

Answer: (B)

$\frac {45}{49}$

Answer 2

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) $\frac {45}{49}$