Q79 / 125 · BITSAT 2009 · Mathematics
The random variable \(\mathrm{X}\) has the following probability distribution
\(\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{x} & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{k} & 3 \mathrm{k} & 5 \mathrm{k} & 2 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}\)
Then the value of \(P(X \geq 2)\) is
- A\(\frac{1}{3}\)
- Correct answerB\(\frac{2}{3}\)
- C\(\frac{3}{4}\)
- D\(\frac{1}{4}\)
Solution
Since, \(\Sigma \mathrm{P}_{\mathrm{i}}(\mathrm{X}=\mathrm{x})=1\)
\(\begin{aligned}
& \therefore \mathrm{k}+3 \mathrm{k}+5 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=1 \therefore 12 \mathrm{k}=1 \therefore \mathrm{k}=\frac{1}{12} \\
& \text { Now, } \mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \\
& =5 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=8 \mathrm{k}=8\left(\frac{1}{12}\right)=\frac{2}{3}
\end{aligned}\)
Hence, the correct answer is option: 'B'.