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If X is a random variable with probability distribution as given below: 

X 0 1 2 3
PX k 3k 3k k

Then, the variance of X is

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Important Questions on Probability Distribution

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IMPORTANT

A random variable X has the following probability distribution:

X=xi 1 2 3 4
PX=xi 0·1 0·2 0·3 0·4

The mean and the standard deviation are, respectively

MEDIUM
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The probability mass function of a random variable X is Px=3-x10, x=-1, 0, 1, 20,otherwise. The value of EX is
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The probability mass function of a random variate X is Px=kx2,x=1, 2, 3, 40,otherwise. Then, E(X)=
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If the probability mass function of a discrete random variable X is Px=Cx3, x=1, 2, 30,otherwise. Then E(X)=
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The probability distribution of the random variable X is given by 
X=x 0 1 3 5
P(X=x) 0.2 0.5 0.2 0.1

Then, 4EX2-Var(X) is equal to

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A random variable X has the following probability mass function
X=x 0 1 2
P(X=x) q2 2pq p2

If p+q=1, then the variance of X is

EASY
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The probability mass function of a random variable X is given by
X=x 0 1 2 3
P(X=x) q3 3q2p 3qp2 p3
If p+q=1, then E(X)=
EASY
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If three fair coins are tossed, where X=number of heads obtained, then E(X) is