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Let X denotes the number of hours a person uses a mobile and the probability distribution of X is as

X01234
P(X)0.1K2K2KK

Then the value of K is

Solution

Correct Option: 1

For any probability distribution, all probabilities must add up to 1.

From the table:

P(X=0)=0.1P(X = 0) = 0.1

P(X=1)=KP(X = 1) = K

P(X=2)=2KP(X = 2) = 2K

P(X=3)=2KP(X = 3) = 2K

P(X=4)=KP(X = 4) = K


Since all probabilities must add up to 1:

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


Substituting the values:

0.1+K+2K+2K+K=10.1 + K + 2K + 2K + K = 1


Combining like terms:

K+2K+2K+K=6KK + 2K + 2K + K = 6K

0.1+6K=10.1 + 6K = 1


Solving for K:

6K=10.16K = 1 - 0.1

6K=0.96K = 0.9

K=0.96K = \dfrac{0.9}{6}

K=0.15K = 0.15


Therefore, the value of K is 0.150.15.

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