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AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

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Question : 40

Total: 160

A random variable X has the following distribution Values of

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

Then P(0<X<6)=

9
10
(
9
10
)2
(
3
10
)
1
10

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