A discrete random variable x has following probability distribution-P x - - k - 2 k - 2 k - 3 k - k - 2 - 2 k - 2 - 7 k - 2- k Then k -A- 1B- 1 - 10C- 9 - 10D- 1 - 2

The correct option is B 1/10 K + 2K + 2K + 3K + K^2 + 2K^2 + 7K^2 + K = 1 ⇒ 10K2 + 9K 1 = 0 ⇒ K = 1/10 ...

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Standard XII

Mathematics

Conditional Probability

A discrete ra...

Question

A discrete random variable x has following probability distribution

$x : 11234567 P (x) : k 2 k 2 k 3 k k^{2} 2 k^{2} 7 k^{2} + k$

Then k=

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Solution

The correct option is B
1/10

$K + 2 K + 2 K + 3 K + K^{2} + 2 K^{2} + 7 K^{2} + K = 1 \Rightarrow 10 K 2 + 9 K - 1 = 0 \Rightarrow K = \frac{1}{10}$

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Conditional Probability

Standard XII Mathematics

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A discrete random variable x has following probability distribution x:1 1 2 3 4 5 6 7P(x): k 2k 2k 3k k2 2k2 7k2+k Then k=

The correct option is B 1/10 K+2K+2K+3K+K2+2K2+7K2+K=1⇒10K2+9K−1=0⇒K=110

A discrete random variable x has following probability distribution

$x : 11234567 P (x) : k 2 k 2 k 3 k k^{2} 2 k^{2} 7 k^{2} + k$

Then k=

The correct option is B
1/10

$K + 2 K + 2 K + 3 K + K^{2} + 2 K^{2} + 7 K^{2} + K = 1 \Rightarrow 10 K 2 + 9 K - 1 = 0 \Rightarrow K = \frac{1}{10}$