Question

The following is the probability distribution of a discrete random variable. If its mean is \(2.81\), then \(\alpha\beta=\) \[ \begin{array}{|c|c|c|c|c|c|} \hline X=x & -1 & -2 & 1 & \alpha & 3\\ \hline P(X=x) & \frac{1}{25} & \beta & \frac14 & \frac{9}{25} & \frac{3}{10}\\ \hline \end{array} \]

• A. \(\frac{1}{10}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{3}{10}\)
• D. \(\frac{1}{25}\)

Hint:

For probability distribution questions, first use \(\sum P(X=x)=1\) to find unknown probabilities. Then use the mean formula \(E(X)=\sum xP(X=x)\) to determine the remaining unknowns.

Answer 1

The Correct Option is B

Concept: For a probability distribution, \[ \sum P(X=x)=1 \] and \[ E(X)=\sum xP(X=x). \] Use these two conditions to find \(\beta\) and \(\alpha\).

Step 1: Find \(\beta\) using \(\sum P(X=x)=1\). \[ \frac{1}{25}+\beta+\frac14+\frac{9}{25}+\frac{3}{10}=1. \] \[ \beta+\frac{1+9}{25}+\frac14+\frac{3}{10}=1. \] \[ \beta+\frac{10}{25}+\frac14+\frac{3}{10}=1. \] \[ \beta+\frac25+\frac14+\frac{3}{10}=1. \] Taking LCM \(20\), \[ \beta+\frac{8+5+6}{20}=1. \] \[ \beta+\frac{19}{20}=1. \] \[ \beta=\frac{1}{20}. \]

Step 2: Use the mean \(E(X)=2.81\). Given, \[ E(X)=2.81=\frac{281}{100}. \] Therefore, \[ (-1)\left(\frac1{25}\right) +(-2)\left(\frac1{20}\right) +(1)\left(\frac14\right) +\alpha\left(\frac9{25}\right) +3\left(\frac3{10}\right) = \frac{281}{100}. \] \[ -\frac1{25}-\frac1{10}+\frac14+\frac{9\alpha}{25}+\frac9{10} = \frac{281}{100}. \] Combining the constant terms, \[ -\frac4{100}-\frac{10}{100}+\frac{25}{100}+\frac{90}{100} +\frac{9\alpha}{25} = \frac{281}{100}. \] \[ \frac{101}{100} +\frac{9\alpha}{25} = \frac{281}{100}. \] \[ \frac{9\alpha}{25} = \frac{180}{100} = \frac95. \] \[ 9\alpha=45. \] \[ \alpha=5. \]

Step 3: Find \(\alpha\beta\). \[ \alpha\beta = 5\left(\frac1{20}\right) = \frac14. \] Therefore, \[ \boxed{\alpha\beta=\frac14} \] \[ \boxed{\text{Answer = (B)}} \]