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MHT CET · Maths · Probability

\(A, B, C\) are three events, one of which must and only one can happen the odds in favour of \(A\) are \(4: 6\), odds against \(B\) are \(7: 3\), then odds against \(C\) are

  1. A 7:3
  2. B 3:7
  3. C 6:4
  4. D 4:6
Verified Solution

Answer & Solution

Correct Answer

(A) 7:3

Step-by-step Solution

Detailed explanation

\(P(A)=\frac{4}{4+6}=\frac{2}{5}, P(B)=\frac{3}{3+7}=\frac{3}{10}\)
\(\because A, B\) and \(C\) are mutually exclusive and exhaustive
\(\begin{aligned}
& \Rightarrow P(A)+P(B)+P(C)=1 \\
& \Rightarrow \frac{2}{5}+\frac{3}{10}+P(C)=1 \\
& \Rightarrow P(C)=\frac{3}{10}
\end{aligned}\)
Now, odds against to the event \(C=\frac{1-P(C)}{P(C)}=\frac{1-\frac{3}{10}}{\frac{3}{10}}=\frac{7}{3}\)
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