Random Variable Problem Using E X and E X Square

❓ Question

Let a random variable $X$ take values $0,1,2,3$ with

$$P(X=0)=p,P(X=1)=p,P(X=2)=P(X=3)$$

and it is given that $\;E(X^2)=2E(X).$
Find the value of $8p-1.$

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🖼️ Question Image

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✍️ Short Solution

1. Let $q = P(X=2)=P(X=3)$. Since total probability is 1:

$$2p+2q=1\Rightarrow p+q=\frac{1}{2}\Rightarrow q=\frac{1}{2}-p\mathrm{.}$$

2. Compute the expectations:

$$E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q.$$
$$E(X^2) = 0^2\cdot p + 1^2\cdot p + 2^2\cdot q + 3^2\cdot q = p + 13q.$$

3. Use the given relation $E(X^2)=2E(X)$:

$$p+13q=2(p+5q)\mathrm{.}$$

Simplify:

$$p+13q=2p+10q\Rightarrow -p+3q=0\Rightarrow 3q=p\mathrm{.}$$

4. Substitute $q = \tfrac12 - p$ into $3q = p$:

$$3\left(\tfrac12 - p\right) = p \quad\Rightarrow\quad \tfrac32 - 3p = p$$
$$\tfrac32 = 4p \quad\Rightarrow\quad p = \tfrac{3}{8}.$$

(Then $q = \tfrac12 - \tfrac{3}{8} = \tfrac{1}{8}$. Check: $3q = 3\cdot\tfrac{1}{8}=\tfrac{3}{8}=p$ — consistent.)

5. Finally compute the required expression:

$$8p - 1 = 8\cdot\frac{3}{8} - 1 = 3 - 1 = 2.$$

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🖼️ Image Solution

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✅ Conclusion & Video Solution

We found $p=\dfrac{3}{8}$ which gives

$$\boxed{{\displaystyle 8p-1=2}}\mathrm{.}$$

So the value of $8p-1$ is 2.