Question

If the probability function of a random variable X is defined by P(X = k) = a`((k + 1)/2^k)` for k - 0, 1, 2, 3, 4, 5, then the probability that X takes a prime value is ______

Options

• `13/20`

• `23/60`

• `11/20`

• `19/60`

Answer 1

If the probability function of a random variable X is defined by P(X = k) = a`((k + 1)/2^k)` for k - 0, 1, 2, 3, 4, 5, then the probability that X takes a prime value is `underline(23/60)`.

Explanation:

X = k	0	1	2	3	4	5
P(X = k)	a	a	`(3a)/4`	`(4a)/8`	`(5a)/16`	`(6a)/32`

Since `sum_{k = 0}^5 p(X = k) = 1,`

`a + a + (3a)/4 + (4a)/8 + (5a)/16 + (6a)/32 = 1`

⇒ `15/4a = 1` ⇒ a = `4/15`

Now, P(X = prime value)

= P(X = 2) + P(X = 3) + P(X = 5)

`(3a)/4 + (4a)/8 + (6a)/32`

= `(23a)/16`

= `23/16 xx 4/15 = 23/60`