JEE MAIN - Mathematics (2008 - No. 5)
It is given that the events $$A$$ and $$B$$ are such that
$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is :
$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is :
$${1 \over 6}$$
$${1 \over 3}$$
$${2 \over 3}$$
$${1 \over 2}$$
Explanation
Given that,
$$P\left( {{A \over B}} \right) = {1 \over 2}$$
$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$$ = $${1 \over 2}$$.............. equation (1)
$$P\left( {{B \over A}} \right) = {2 \over 3}$$
$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over {P\left( A \right)}}$$ = $${2 \over 3}$$.............. equation (2)
Dividing equation (1) by equation (2) we get,
$${{P\left( A \right)} \over {P\left( B \right)}}$$ = $${3 \over 4}$$
$$ \Rightarrow $$ $${P\left( B \right)}$$ = $${4 \over 3}$$ $$ \times $$ $${P\left( A \right)}$$
= $${4 \over 3}$$ $$ \times $$ $${1 \over 4}$$
= $${1 \over 3}$$
$$\therefore$$ Option (B) is correct.
$$P\left( {{A \over B}} \right) = {1 \over 2}$$
$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$$ = $${1 \over 2}$$.............. equation (1)
$$P\left( {{B \over A}} \right) = {2 \over 3}$$
$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over {P\left( A \right)}}$$ = $${2 \over 3}$$.............. equation (2)
Dividing equation (1) by equation (2) we get,
$${{P\left( A \right)} \over {P\left( B \right)}}$$ = $${3 \over 4}$$
$$ \Rightarrow $$ $${P\left( B \right)}$$ = $${4 \over 3}$$ $$ \times $$ $${P\left( A \right)}$$
= $${4 \over 3}$$ $$ \times $$ $${1 \over 4}$$
= $${1 \over 3}$$
$$\therefore$$ Option (B) is correct.
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