If a random variable X has the following probability distribution values of X
Then `P(Xge6)` is equal to

Options

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

Answer

A

Explanation

Since , the sum of all the probabillities in a probability distribution is always unity
`therefore" "P(X=0)+P(X=1)+...+P(X=7)=1`
`rArr0+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" "[therefore" probability can not be negative"]`
Now, `P(Xge6)=P(X=6)+(X=7)`
`=2k^2+7k^2+k=9k^2+k=9/(100)+1/(10)=(19)/(100)" "[thereforek=1//10]`