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CUET
-- Mathematics - Section A
Probability
Probability distribution of random variable X is
X
0
1
2
3
4
5
P(X)
3k
2k
2k2
3k2
4k2
The value of k is:
$k=-1, \frac{1}{9}$
$k=-1$
$k=\frac{1}{9}$
$k=0,-9$
The correct answer is Option (3) - $k=\frac{1}{9}$