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The probability distribution of a random discrete variable is given

X-10123
P(X)0.1pp0.3qqrr

If it is known that P(X=1) is the mean of P(X=0) and P(X=2).

Then the value of r is :

Solution

✅ Correct Option: 1

The probability distribution has some missing values:

X-10123
P(X)0.1pp0.3qqrr

P(X=1) is the mean of P(X=0) and P(X=2).


P(X=1) is the mean of P(X=0) and P(X=2):

0.3=p+q20.3 = \dfrac{p + q}{2}

p+q=0.6p + q = 0.6


All probabilities in a probability distribution must sum to 1:

0.1+p+0.3+q+r=10.1 + p + 0.3 + q + r = 1

0.4+p+q+r=10.4 + p + q + r = 1

p+q+r=0.6p + q + r = 0.6


From the equations:

p+q=0.6p + q = 0.6

p+q+r=0.6p + q + r = 0.6

Therefore:

0.6+r=0.60.6 + r = 0.6

r=0r = 0

The value of r=0r = 0.

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