A random variable X has the following probability distribution- begin array c - c - c - c - c - c - c - c - c P X - 0 - k - 2 k - 2 k - 3 k - k - 2 - 2 k - 2 - 7 k - 2- k Whlinelendarray P X -6

P(X gt;6)=P(7)=7k^2+k=7/100+1/10=17/100 (Put k=1/10) ...

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Standard XII

Mathematics

Intersection of Sets

A random vari...

Question

A random variable X has the following probability distribution.

$\begin{matrix} X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 P (X) & 0 & k & 2 k & 2 k & 3 k & k^{2} & 2 k^{2} & 7 k^{2} + k \end{matrix}$

P(X > 6)

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Solution

$P (X > 6) = P (7) = 7 k^{2} + k = \frac{7}{100} + \frac{1}{10} = \frac{17}{100}$ $(P u t k = \frac{1}{10})$

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A random variable X has the following probability distribution. X 0 1 2 3 4 5 6 7P(X)0 k 2k 2k 3k k2 2k2 7k2+k P(X > 6)

P(X>6)=P(7)=7k2+k=7100+110=17100 (Put k=110)

A random variable X has the following probability distribution.

$\begin{matrix} X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 P (X) & 0 & k & 2 k & 2 k & 3 k & k^{2} & 2 k^{2} & 7 k^{2} + k \end{matrix}$

P(X > 6)

$P (X > 6) = P (7) = 7 k^{2} + k = \frac{7}{100} + \frac{1}{10} = \frac{17}{100}$ $(P u t k = \frac{1}{10})$