find the value of p as the random variable X =0 1 2 3 given p (x=0) =p (x=1)=p and p (x=2)=p (x=3)
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Since X can take values only from {0, 1, 2, 3}, P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1, as it includes the probabilities of all the possible events.
Let P(X = 2) = P(X = 3) = k.
Then p + p + k + k = 1
:: p + k = 1/2
Now, sum of all pX^2 = 2 sum of all px
:: p(0)^2 + p(1)^2 + k(2)^2 + k(3)^2 = 2[p(0) + p(1) + k(2) + k(3)]
:: p + 13k = 2p + 10k
:: p = 3k
:: k = p/3
Substituting this value of k in the 1st equation, we have:
p + p/3 = 1/2
:: p = 3/8, which is the required answer.
Let P(X = 2) = P(X = 3) = k.
Then p + p + k + k = 1
:: p + k = 1/2
Now, sum of all pX^2 = 2 sum of all px
:: p(0)^2 + p(1)^2 + k(2)^2 + k(3)^2 = 2[p(0) + p(1) + k(2) + k(3)]
:: p + 13k = 2p + 10k
:: p = 3k
:: k = p/3
Substituting this value of k in the 1st equation, we have:
p + p/3 = 1/2
:: p = 3/8, which is the required answer.
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