Let $X$ be a discrete random variable. The probability distribution of $X$ is given below $…

Let $X$ be a discrete random variable. The probability distribution of $X$ is given below $ \begin{array}{|l|c|c|c|} \hline X & 30 & 10 & -10 \\ \hline P(X) & \frac{1}{5} & A & B \\ \hline \end{array} $ and $E(X)=4$, then the value of $AB$ is equal to.
  1. $\frac{3}{10}$
  2. $\frac{2}{15}$
  3. $\frac{1}{15}$
  4. $\frac{3}{20}$

Solution

The probability distribution is defined with $A + B = \frac{4}{5}$ from the total probability constraint.

Given $E(X) = 4$, apply the expectation formula: $4 = 30 \cdot \frac{1}{5} + 10A - 10B$, simplifying to $5A - 5B = -1$.

Multiply the total probability equation by 5: $5A + 5B = 4$. Adding this to $5A - 5B = -1$ yields $10A = 3$, so $A = \frac{3}{10}$.

Substitute into $A + B = \frac{4}{5}$: $B = \frac{4}{5} - \frac{3}{10} = \frac{1}{2}$.

Then $AB = \frac{3}{10} \cdot \frac{1}{2} = \frac{3}{20}$, matching option D.

Asked in: MHT CET 2025 (19 April Shift 2)

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