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

Options

A. `(19)/(100)`
B. `(81)/(100)`
C. `(9)/(100)`
D. `(91)/(100)`

Answer

A

Explanation

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]`