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The discrete random variable X has the probability function. Valueof X = x 0 1 2 3 4 5 6 7 P(x) 0 k 2k 2k 3k k2 2k2 7k2 + k Find k

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प्रश्न

The discrete random variable X has the probability function.

Value
of X = x
0 1 2 3 4 5 6 7
P(x) 0 k 2k 2k 3k k2 2k2 7k2 + k

Find k

योग
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उत्तर

Since the condition of Probability Mass function

`sum_("i" = 2)^oo` P(xi) = 1

`sum_("i" = 0)^7` P(xi) = 1

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

P(x = 6) + P(x = 7) = 1

0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1

10k2 + 9k – 1 = 0

10k2 + 10k – k – 1 = 0

10k(k + 1) – 1(k + 1) = 0

(k + 1)(10k – 1) = 0

k + 1 = 0, 10k – 1 = 0

k = – 1, 10k = 1

k = `1/10`

k = – 1 is not possible

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Random Variable and Mathematical expectation - Exercise 6.1 [पृष्ठ १३३]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 6 Random Variable and Mathematical expectation
Exercise 6.1 | Q 6. (i) | पृष्ठ १३३

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