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I posted a letter to my friend and do not receive a reply. It is known that one out of l0 letters do not reach the destination. It is certain that my friend would have replied if he received the letter. Denotes the event that my friend receives the letter and B that I get a reply then, which of the following is/ are correct

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$P (A \cap B) = \frac{81}{100}$

$P (A \cup B) = \frac{9}{20}$

$P (\frac{A}{B^{1}}) = \frac{9}{19}$

$P (B) = \frac{81}{100}$

answer is A, B, C.

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Detailed Solution

$\begin{aligned} P (A) = \frac{9}{10}, P (A^{'}) = \frac{1}{10} \\ P (B ∣ A) = \frac{9}{10}, P (B ∣ A^{'}) = 0 \end{aligned}$

$Now P (B ∣ A) = \frac{9}{10} \Rightarrow \frac{P (A \cap B)}{P (A)} = \frac{9}{10}$

$\Rightarrow P (A \cap B) = {(\frac{9}{10})}^{2}$
$P (B) = P (A) P (\frac{B}{A}) + P (A^{1}) P = (\frac{B}{A^{1}})$

$= (\frac{9}{10}) (\frac{9}{10}) + (\frac{1}{10}) (0) = {(\frac{9}{10})}^{2}$

$\Rightarrow P (B^{'}) = 1 - P (B) = 1 - {(\frac{9}{10})}^{2} = \frac{19}{100}$

$∴ P (A ∣ B^{'}) = \frac{P (A \cap B)}{P (B^{'})} = \frac{P (A) - P (A \cap B)}{P (B^{'})} = \frac{9}{19}$

$P (A \cup B) = \frac{9}{10} + {(\frac{9}{10})}^{2} - {(\frac{9}{10})}^{2} = \frac{9}{10}$

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