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  • A random variable X has the following probability distribution:          <img alt="X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5" src="https://entrancecorner.oncodecogs.com/gif.latex

A random variable X has the following probability distribution:          X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5 P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2} Then P(X>2) is equal to: 
Option: 1 \frac{7}{12}
Option: 2 \frac{23}{36}
Option: 3 \frac{1}{36}
Option: 4 \frac{1}{6}
 

Answers (11)

best_answer

\\\sum P_i=1\Rightarrow 6k^2+5k=1\\\\\Rightarrow 6k^2+5k-1=0\\\\\Rightarrow k=\frac{1}{6},\:k=-1\text{ (invalid)}\\\\ Now,\,\,{P(X>2)=P(3)+P(4)+P(5)=k+2 k+5 k^{2}} \\\\ {=\frac{1}{6}+\frac{2}{6}+\frac{5}{36}=\frac{6+12+5}{36}=\frac{23}{36}}

Correct Option 2

Posted by

avinash.dongre

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 Correct option 2

Posted by

Deepak ved

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23/36

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Megha Bopche

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23/36

Posted by

Brijesh kanav

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1

Posted by

Kamlesh kumar

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7/12

Posted by

Kamlesh kumar

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2

Posted by

Surjeet Kumar

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option 2

 

Posted by

Rohit Mangal

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option 2 

Harshita Mangal

 

Posted by

Rohit Mangal

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Option 2

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ritik.gaur

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Option 2

Posted by

kpvalli2725@gmail.com

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