Question

The p.m.f. of a r.v. X is as follows:
P(X = 1) = 5k³, P(X = 2) = 10k – 16k² – 1
P(X = 3) = 2k, P(X = 4) = k, P(X = x) = 0 for any other value of x. Then the value of k is ______.

पर्याय

• `1/5`

• `4/5`

• 1

• 5

Answer 1

The p.m.f. of a r.v. X is as follows:
P(X = 1) = 5k³, P(X = 2) = 10k – 16k² – 1
P(X = 3) = 2k, P(X = 4) = k, P(X = x) = 0 for any other value of x. Then the value of k is `1/5`.

Explanation:

Since, `sum_(x = 1)^4 "P"("X" = x)` = 1

∴ 5k³ + 10k – 16k² – 1 + 2k  + k = 1

⇒ 5k³ – 16k² + 13k – 2 = 0

⇒ (k – 1)(k – 2)(5k – 1) = 0

⇒ k = 1 or k = 2 or k = `1/5`

For k = 1 or k = 2, P(X = 1) > 1, which is not possible.

∴ k = `1/5`