Question -
Answer -
Given: A randomvariable X with its probability distribution.
(a) As we know the sumof all the probabilities in a probability distribution of a random variablemust be one.
Hence the sum ofprobabilities of given table:
⇒ k + 2k + 3k + 0= 1
⇒ 6k = 1
k = 1/6
(b) Now we have tofind P(X < 2)
P (X < 2) = P (X =0) + P (X = 1)
= k + 2k
= 3k
P (X < 2) = 3 × 1/6= ½
Consider P (X ≤ 2)
P (X ≤ 2) = P (X = 0)+ P (X = 1) + P (X = 2)
= k + 2k + 3k
= 6k
P (X ≤ 2) = 6 × 1/6 =1
Now we have to findP(X ≥ 2)
P(X ≥ 2) = P(X = 2) +P(X > 2)
= 3k + 0
= 3k
P (X ≥ 2) = 3 × 1/6 =½