Find:

(i) P(X<4)

(ii) P (3≤ X≤6)

Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

Note:- Sum of all probabilities = 1

Therefore, k + 3k + 5k + 7k + 9k + 11k + 13k = 1

$ \therefore k = \frac{1}{49} $

(i) P(X<4) = P(0) + P(1) + P(2) + P(3)

$ = \frac{1}{49} + \frac{3}{49} + \frac{5}{49} + \frac{7}{49} = \frac{16}{49}$

(ii) P(3 $\leq$ X $\leq$ 6) = P(3) + P(4) + P(5) + P(6)

$ = \frac{7}{49} + \frac{9}{49} + \frac{11}{49} + \frac{13}{49} = \frac{40}{49} $