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Tamil Nadu Board of Secondary EducationHSC Science Class 12

A random variable X has the following probability mass function. x 1 2 3 4 5 F(x) k2 2k2 3k2 2k 3k Find the value of k

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Question

A random variable X has the following probability mass function.

x 1 2 3 4 5
F(x) k2 2k2 3k2 2k 3k

Find the value of k

Chart
Sum
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Solution

Given f(x) in a probability mass function

`sum_x` f(x) = 1

k2 + 2k2 + 3k2 + 2k + 3k = 1

6k2 + 5k = 1

6k2 + 5k – 1 = 0

(k + 1)(6k – 1) = 0

k = `1/6`

k ≠ –1 neglecting negative terms

Probability mass function

x 1 2 3 4 5
F(x) `1/36` `2/36` `3/36` `2/6` `3/6`
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Chapter 11: Probability Distributions - Exercise 11.2 [Page 194]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 11 Probability Distributions
Exercise 11.2 | Q 6. (i) | Page 194

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