MHT CET · Maths · Probability
A random variable \(X\) has the following. -probability distribution
Then \(\mathrm{P}(\mathrm{X}\gt2)\) is equal to
- A \(\frac{7}{12}\)
- B \(\frac{23}{36}\)
- C \(\frac{1}{36}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{23}{36}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{k}^2+2 \mathrm{k}+\mathrm{k}+2 \mathrm{k}+5 \mathrm{k}^2=1 \\
& \therefore \quad 6 \mathrm{k}^2+5 \mathrm{k}=1 \\
& \therefore \quad 6 \mathrm{k}^2+5 \mathrm{k}-1=0 \\
& \therefore \quad 6 \mathrm{k}^2+6 \mathrm{k}-\mathrm{k}-1=0 \\
& \therefore \quad(6 k-1)(k+1)=0 \\
& \therefore \quad k=\frac{1}{6} \quad \ldots[\because k=-1 \text { is not possible }] \\
& \therefore \quad P(X\gt2)=1-P(X \leq 2) \\
& =1-\left(\mathrm{k}^2+2 \mathrm{k}\right) \\
& =1-\left(\frac{1}{36}+\frac{2}{6}\right) \\
& =1-\frac{13}{36} \\
& =\frac{23}{36}
\end{aligned}\)
& \mathrm{k}^2+2 \mathrm{k}+\mathrm{k}+2 \mathrm{k}+5 \mathrm{k}^2=1 \\
& \therefore \quad 6 \mathrm{k}^2+5 \mathrm{k}=1 \\
& \therefore \quad 6 \mathrm{k}^2+5 \mathrm{k}-1=0 \\
& \therefore \quad 6 \mathrm{k}^2+6 \mathrm{k}-\mathrm{k}-1=0 \\
& \therefore \quad(6 k-1)(k+1)=0 \\
& \therefore \quad k=\frac{1}{6} \quad \ldots[\because k=-1 \text { is not possible }] \\
& \therefore \quad P(X\gt2)=1-P(X \leq 2) \\
& =1-\left(\mathrm{k}^2+2 \mathrm{k}\right) \\
& =1-\left(\frac{1}{36}+\frac{2}{6}\right) \\
& =1-\frac{13}{36} \\
& =\frac{23}{36}
\end{aligned}\)
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