Question

A random, variable `X` has the following probability distribution: `X : 0 1 2 3 4 5 6 7` `P(X):0 k 2k 2k 3k k^2 2k^2 7k^2+k` Find each of the following: `k` ii. `P(X<6)` iii. `P(|Xgeq6)` iv. `P(0<x ,5)`<br>A. `(19)/(100)` B. `(81)/(100)` C. `(9)/(100)` D. `(91)/(100)` Select the correct answer from above options

Answer 1

Correct Answer - A Since the sum of the probabilities in a probability distribution is always unity. `therefore P(X=0)+P(X=1)+.....+P(X=7)=1` `rArr 0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1` `rArr 10k^2+9k-1=0` `rArr (10k-1)(k+1)=0` `rArr 10k-1=0 " "[because kge0thereforek+1 ne 0]` `rArr k=(1)/(10)` Now, `P(Xge 6)=P(X=6)+P(X=7)=2k^2+7k^2+k=9k^2+k=(19)/(100)" "(k=1//10]`