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Let X be a discrete random variable whose probability distribution is defined as follows:P(X = x) = kfor,,,kfor,,Otherwise{k(x+1) for x=1,2,3,42kx for x=5,6,70 - Mathematics

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Question

Let X be a discrete random variable whose probability distribution is defined as follows:
P(X = x) = `{{:("k"(x + 1),  "for"  x = 1"," 2"," 3"," 4),(2"k"x,  "for"  x = 5"," 6"," 7),(0,  "Otherwise"):}`
where k is a constant. Calculate the value of k

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Solution

Here, P(X = x) = k(x + 1) for x = 1, 2, 3, 4

So, P(X = 1) = k(1 + 1) = 2k

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

P(X = 3) = k(3 + 1) = 4k

P(X = 4) = k(4 + 1) = 5k

Also, P(X = x) = 2kx for x = 5, 6, 7

P(X = 5) = 2(5)k = 10k

P(X = 6) = 2(6)k = 12k

P(X = 7) = 2(7)k = 14k

And for otherwise it is 0.

∴ The probability distribution is given by

X 1 2 3 4 5 6 7 Otherwise
P(X) 2k 3k 4k 5k 10k 12k 14k 0

We know that `sum_("i" = 1)^"n" "P"("X"_"i")` = 1

So, 2k + 3k + 4k + 5k + 10k + 12k + 14k = 1

⇒ 50k = 1

⇒ k = `1/50`

Hence, the value of k is `1/50`

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Chapter 13: Probability - Exercise [Page 277]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 13 Probability
Exercise | Q 49.(i) | Page 277

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