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Exercise 11.6 · Q11

Q.If P(X=0)=1−P(X=1)P(X=0)=1-P(X=1) and E(X)=3 Var(X)E(X)=3\,\text{Var}(X), then P(X=0)P(X=0) is

(1) 23\dfrac23
(2) 25\dfrac25
(3) 15\dfrac15
(4) 13\dfrac13.
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P(X=0)=1−P(X=1)P(X=0)=1-P(X=1) means XX only takes the values 0,10,1 (a Bernoulli-shaped variable), so its mean is p=P(X=1)p=P(X=1) and its variance is pqpq; substitute both into the given relation E(X)=3 Var(X)E(X)=3\,\text{Var}(X).

Step 1. Set up notation. Let p=P(X=1)p=P(X=1), so P(X=0)=1−p=qP(X=0)=1-p=q.

Step 2. Mean and variance of this two-valued XX. E(X)=0⋅q+1⋅p=pE(X)=0\cdot q+1\cdot p=p. V(X)=E(X2)−(E(X))2=(02q+12p)−p2=p−p2=p(1−p)=pqV(X)=E(X^2)-(E(X))^2=(0^2q+1^2p)-p^2=p-p^2=p(1-p)=pq. …

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