TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Section: Mathematics
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Question : 41 of 160
 
Marks: +1, -0
The probability distribution of a random variable X is given below:
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<4)=
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