# Choose the correct option from the given alternative: If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 −x , where k is a constant, then P (X = 0) = - Mathematics and Statistics

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Choose the correct option from the given alternative:

If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 −x , where k is a constant, then P (X = 0) =

#### Options

• 7/25

• 16/25

• 18/25

• 19/25

#### Solution

If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 −x , where k is a constant, then P (X = 0) = 16/25

Hint : k[1/5^0 + 2/5^1 + 3/5^2 + ....]=1

Let s = k/5^0+ 2k/5^1+3k/5^2+ ...............

i.e s = k + 2k/5+3k/5^2+..........

∴1/5s = k/5 + (2k)/5^2 + (3k)/5^3 + ........

∴ s - 1/5s = k + k/5 + k/5^2 + k/5^3 + ..........

∴ 4/5s=k[1 + 1/5 + 1/5^2 + 1/5^3 + ....]

= k [1/(1 - 1/5)] = 5k/4

∴ s = (25k)/16 = 1

∴ k = 16/25

∴P(x=0) = k(0+1)5^0 = k = 16/25.

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise | Q 4 | Page 242