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The probability distribution of discrete r.v. X is as follows : x = x 1 2 3 4 5 6 P[x=x] k 2k 3k 4k 5k 6k (i) Determine the value of k. (ii) Find P(X≤4), P(2<X< 4), P(X 3). - Mathematics and Statistics

Sum

The probability distribution of discrete r.v. X is as follows :

x = x 1 2 3 4 5 6
P[x=x] k 2k 3k 4k 5k 6k

(i) Determine the value of k.

(ii) Find P(X≤4), P(2<X< 4), P(X≥3).

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Solution

(i) Since P (x) is a probability distribution of x,

`Sigma_(x = 0)^6 P(x) = 1`

∴ P (0) + P (1) + P(2) + P(3) + P(4) + P(5) + P (6) = 1

∴ k + 2k + 3k + 4k + 5k + 6k = 1

∴ 21k = 1

∴ k = `1/21`

(ii)

P(X ≤ 4)

= P (0) + P (1) + P(2) + P(3) + P(4)

=k + 2k + 3k + 4k

= 10k

= `10 (1/21)`

= `10/21`

 P(2< X < 4)

= P(3)

= 3k

= `3(1/21)`

= `1/7`

P(X ≥ 3)

= P (1) + P(2) + P(3)

= k +  2k + 3k 

= 6k

= `6(1/21)`

= `2/7`

  Is there an error in this question or solution?
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APPEARS IN

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