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A random variable X has the following probability distribution

X012345
P(X)0.1k0.22k0.3k

Then P (X < 3) is

Solution

Correct Option: 4

A probability distribution table is given with some known values and some unknown values in terms of kk.

In any probability distribution, all probabilities must add up to 11.


Adding all the probabilities and setting them equal to 1:

P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=1P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1

0.1+k+0.2+2k+0.3+k=10.1 + k + 0.2 + 2k + 0.3 + k = 1

0.6+4k=10.6 + 4k = 1

4k=0.44k = 0.4

k=0.1k = 0.1


P(X<3)P(X < 3) means the probability that XX is less than 3.

This includes X=0,1,2X = 0, 1, 2 (not 3).


P(X<3)=P(X=0)+P(X=1)+P(X=2)P(X < 3) = P(X=0) + P(X=1) + P(X=2)

P(X<3)=0.1+k+0.2P(X < 3) = 0.1 + k + 0.2

P(X<3)=0.1+0.1+0.2P(X < 3) = 0.1 + 0.1 + 0.2

P(X<3)=0.4P(X < 3) = 0.4

Therefore, P(X<3)=0.4P(X < 3) = 0.4

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