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Simple Pendulum

Given two eve...

Question

Given two events $A$ and $B$ . If odds against $A$ are $2 : 1$ and those in favour of $A \cup B$ are $3 : 1$ , then

\frac{1}{2} \leq P (B) \leq \frac{3}{4}

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\frac{5}{12} \leq P (B) \leq \frac{3}{4}

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\frac{1}{4} \leq P (B) \leq \frac{3}{5}

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None of these

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Solution

The correct option is C $\frac{5}{12} \leq P (B) \leq \frac{3}{4}$
Given that, $P (A) = \frac{1}{3}$ , $P (A \cup B) = \frac{3}{4}$
Now, $P (A \cup B) = P (A) + P (B) - P (A \cap B) \leq P (A) + P (B)$
$\Rightarrow \frac{3}{4} \leq \frac{1}{3} + P (B) \Rightarrow \frac{5}{12} \leq P (B)$
and $P (B) \leq P (A \cup B) = \frac{3}{4}$
$∴ \frac{5}{12} \leq P (B) \leq \frac{3}{4}$

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Similar questions

2 : 1

A \cup B

3 : 1

\frac{1}{12} \leq P (B) \leq \frac{3}{4}

A \cap B \subset A

P (A \cap B) \leq P (A)

P (A) = \frac{2}{5}

P (B) = \frac{3}{4}

\frac{1}{20} \leq P (A \cap B) \leq \frac{2}{5}

P (A \cup B) \leq max {P (A), P (B)}

P (A \cap B) \geq min {P (A), P (B)}

P (A) = \frac{1}{2}

P (B) = \frac{1}{3}

P (A \cap B) = \frac{1}{4}

P (B / A)

P (A) = P (A / B) = \frac{1}{4}

P (B / A) = \frac{1}{2}

P (¯ A / B) = \frac{3}{4}

P (¯ B / A) = \frac{1}{2}

P (B / A) = P (B)

If A and B are any two events such that $P (A) = \frac{1}{2}, P (B) = \frac{1}{3}, a n d P (A \cap B) = \frac{1}{4}, t h e n P (A^{'} / B^{'}) =$

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