A random variable `X` has the following probability distribution:

Determine :
(i) `k`
(ii) `P(Xlt3)`
(iii) `P(Xgt6)`
(iv) `P(0ltXlt3)`

Answer

N/a

Explanation

(i) Since the sum of the probabilities of a probability distribution of a random variable is 1.
i.e. `sumP(X)=1`
Therefore `P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)=1`
`implies0+k+2k+2k+3k+k^(2)+2k^(2)+7k^(2)+k=1`
`implies10k^(2)+9k-1=0`
`implies10k^(2)+10k-k-1=0`
`implies10k(k+1)-1(k+1)=0`
`implies(k+1)(10k-1)=0`
`impliesk+1=0` or `10k-1=0`
`impliesk=-1` or `k=1/10`
Because `k=-1` is not possible as the probability of an event can never be negative.
`:.k=1/10`
(ii) `P(Xlt3)=P(0)=P(1)+P(2)=0+k+2k`
`=0+1/10+2/10=3/10`
(iii) `P(Xgt6)=P(7)=7k^(2)+k=7/100+1/10=17/100`
(iv) `P(0ltXlt3)-P(1)+P(2)=k+2k`
`=1/10+2/10=3/10`