Before a race the chance of three runners A, B & C were estimated to be proportional to 5, 3 & 2 respectively but during the race A meets with an accident which reduces his chance to 1/3. If the respective chances of B and C are P(B) and P(C) then

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

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P (C) = \frac{4}{15}

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

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P (B) = \frac{4}{15}

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Solution

The correct options are
A $P (C) = \frac{4}{15}$
B $P (B) = \frac{2}{5}$
Ratio of their probabilities are 5 : 3 : 2
$\Rightarrow 5 x + 3 x + 2 x = 1 \Rightarrow x = \frac{1}{10}$
their respective probabilities of A, B & C will be
$= \frac{1}{2}, \frac{3}{10}, \frac{1}{5}$
Now After A's accident
$\frac{1}{2} - t = \frac{1}{3}$
$\Rightarrow t = \frac{1}{6}$ probability get reduced which will be shared between B & C
$\Rightarrow 3 y + 2 y = \frac{1}{6} \Rightarrow y = \frac{1}{30}$
$\Rightarrow$ Probability of C $= \frac{1}{5} + 2 \times \frac{1}{30} = \frac{4}{15}$
Probability of B $= \frac{3}{10} + 3 \times \frac{1}{30} = \frac{2}{5}$

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B

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Three events A, B and C have probabilities $\frac{2}{5}, \frac{1}{3}$ and $\frac{1}{2}$ , respectively. If,

$P (A \cap C) = \frac{1}{5} a n d P (B \cap C) = \frac{1}{4}$ , then find the values of $P (\frac{C}{B}) a n d P (A^{'} \cap C^{'}) .$

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

P (B) = \frac{1}{6}

P (C) = \frac{2}{3}

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P (A) = \frac{1}{2}, P (B) = \frac{1}{3}, P (C) = \frac{1}{2} a n d

P (A \cap B) = P (B \cap C) = P (C \cap A) = \frac{1}{4} a n d

P (A \cap B \cap C) = \frac{1}{8}

P (A \cup B \cup C)

A, B, C

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) + P (A \cap B \cap C)

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