$Find the odds of event b$

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Given an event has exactly four outcomes, a , b , c , and d , where event a , b , and c are independent events. Given a represents a few outcomes of d , i.e., a is a proper subset of d , find the odds in favour of event b , if P (a | b ) = 1 / 6 , P (a | d ) = 4 / 9 , and 3P (c ) = P (d )

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Ilsrc

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Since $a$ and $b$ are independent \[P(a|b)=P(a)=\frac{1}{6}.\] Also since $a$ is a subset of $b$, \[\frac{4}{9}=P(a|d)=\frac{P(a \cap d)}{p(d)}=\frac{P(a)}{P(d)}=\frac{1/6}{P(d)}\] \[\Rightarrow P(d)=\frac{\frac{1}{6}}{\frac{4}{9}}=\frac{3}{8}.\] \[P(c)=\frac{P(d)}{3}=\frac{\frac{3}{8}}{3}=\frac{1}{8}.\] On the other hand (note that $a$ is a subset of $d$) \[1=P(b \cup c \cup d)=P(b)+P(c)+P(d)-P(b\cap c) - P(b \cap d)-P(c \cap d)+P(b\cap c \cap d)\] \[=P(b)+P(c)+P(d)-P(b)P(c)-P(b)P(d)-P(c)P(d)+P(b)P(c)P(d)\] \[=P(b)+\frac{1}{8}+\frac{3}{8}-P(b) \times \frac{1}{8}-P(b) \times \frac{3}{8}-\frac{1}{8}\times\frac{3}{8}+P(b)\times \frac{1}{8}\times\frac{3}{8}\] \[=P(b)[1-\frac{1}{8}-\frac{3}{8}+\frac{3}{64}]+\frac{4}{8}-\frac{3}{64}=1\] \[\Rightarrow P(b)(\frac{35}{64}) =1-\frac{29}{64}=\frac{35}{64}\]
\[\Rightarrow P(b)=1.\]

Nirenberg

Nirenberg

This was a very difficult question. It took me about 45 minutes to solve it. Please offer higher amounts for such questions in future, otherwise you may not get a response from other users.
Ilsrc

Okay thank you so muchhh
Nirenberg

You're welcome. I am glad I could help.

$Find the odds of event b$

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