A random variable X has the following probability distribution. Then the value of (i) k (ii) `P(Xlt3)` (iii) `P(Xgt6)` (iv) `P(0ltXlt3)`

Options

A. `1/(10),3/(10),(17)/(100) and 3/(10)`
B. `1/(10),3/(10),3/(10) and (17)/(100)`
C. `(17)/(100),1/(10),3/(10) and 3/(10)`
D. None of these

Answer

A

Explanation

It is known that the sum of probabillties of a probability distribution of random variable is one i.e
`SigmaP(X)=1` therefore
`P(0)+P(1)+P(2)+P(4)+P(5)+P(6)+P(7)=1`
`rArr" "=+k+2k+2k+3k+k^2+2k^2+7k^2+k=1`
`rArr" "10k^2+9k-1=0`
`rArr" "10k^2+10k-k-1=0`
`rArr" "10k(k+1)-1(k+1)=0`
`rArr" "(k+1)(10k-1)=0`
`rArr" "k+1=0 or 10k-1=0`
`rArr" "k=-1 or k=1/(10)`
k=-1 is not possible as the probability of an event is never negative
`k=1/(10)`
(ii) `P(Xlt3)=P(0)+P(1)+P(2)=0+k+2k`
`=0+1/(10)+2/(10)=3/(10)" "["put "k=1/(10)]`
(iii) `P(Xgt6)=P(7)=7k^2+k`
`=7/(100)+1/(10)=(17)/(100)" "["put "k=1/(10)]`
(iv) `P(0ltXlt3)=P(1)+P(2)`
`=k+2k=1/(10)+2/(10)=3/(10)" "["put "k=1/(10)]`