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Question

A random variable X has the following probability distribution:
X12345P(X)K22KK2K5K2
Then P(X>2) is equal to:

A
712
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B
2336
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C
136
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D
16
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Solution

The correct option is B 2336
We know that 5x=1P(X)=1
K2+2K+K+2K+5K2=1
K=1,16 K=16

P(X>2)=P(X=3)+P(X=4)+P(X=5)
=K+2K+5K2=2336

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