MHT CET · Maths · Probability
There are three events \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), one of which must and only one can happen. The odds are 8:3 against \(\mathrm{A}, 5: 2\) against B and the odds against C is \(43: 17 \mathrm{k}\), then value of \(k\) is
- A \(\frac{1}{2}\)
- B 2
- C \(\frac{1}{3}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
The odds against A are \(8: 3\).
\(\therefore \quad \mathrm{P}(\mathrm{~A})=\frac{3}{11}\)
Odds against B are \(5: 2\)
\(\begin{array}{ll}
\therefore & P(B)=\frac{2}{7} \\
\therefore & P(A)+P(B)+P(C)=1 \\
& \Rightarrow \frac{3}{11}+\frac{2}{7}+P(C)=1 \\
\therefore & P(C)=1-\frac{2}{7}-\frac{3}{11} \\
\therefore & P(C)=\frac{34}{77}
\end{array}\)
\(\therefore \quad\) odds against \(\mathrm{P}(\mathrm{C})=\frac{77-34}{34}=\frac{43}{34}\)
But odds against \(\mathrm{C}=\frac{43}{17 \mathrm{k}}\)
\(\begin{aligned}
& \therefore \quad \frac{43}{17 k}=\frac{43}{34} \\
& \therefore \quad k=2
\end{aligned}\)
\(\therefore \quad \mathrm{P}(\mathrm{~A})=\frac{3}{11}\)
Odds against B are \(5: 2\)
\(\begin{array}{ll}
\therefore & P(B)=\frac{2}{7} \\
\therefore & P(A)+P(B)+P(C)=1 \\
& \Rightarrow \frac{3}{11}+\frac{2}{7}+P(C)=1 \\
\therefore & P(C)=1-\frac{2}{7}-\frac{3}{11} \\
\therefore & P(C)=\frac{34}{77}
\end{array}\)
\(\therefore \quad\) odds against \(\mathrm{P}(\mathrm{C})=\frac{77-34}{34}=\frac{43}{34}\)
But odds against \(\mathrm{C}=\frac{43}{17 \mathrm{k}}\)
\(\begin{aligned}
& \therefore \quad \frac{43}{17 k}=\frac{43}{34} \\
& \therefore \quad k=2
\end{aligned}\)
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