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If the random variable X has the following probability distribution:

X012Otherwise
P(X)K3K5K0

then K is equal to

Solution

✅ Correct Option: 3

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

From the table, the random variable XX can take values 0, 1, or 2 with probabilities:

P(X=0)=KP(X = 0) = K

P(X=1)=3KP(X = 1) = 3K

P(X=2)=5KP(X = 2) = 5K

P(X=otherwise)=0P(X = \text{otherwise}) = 0


Since all probabilities must sum to 1:

K+3K+5K+0=1K + 3K + 5K + 0 = 1

9K=19K = 1

K=19K = \dfrac{1}{9}


Therefore, K=19K = \dfrac{1}{9}

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