JEE Advance - Mathematics (2017 - Paper 1 Offline - No. 1)
Let X and Y be two events such that $$P(X) = {1 \over 3}$$, $$P(X|Y) = {1 \over 2}$$ and $$P(Y|X) = {2 \over 5}$$. Then
$$P(Y) = {4 \over {15}}$$
$$P(X'|Y) = {1 \over 2}$$
$$P(X \cup Y) = {2 \over 5}$$
$$P(X \cap Y) = {1 \over 5}$$
Explanation
$$P(X) = {1 \over 3}$$
$$P\left( {{X \over Y}} \right) = {{P(X \cap Y)} \over {P(Y)}} = {1 \over 2}$$
$$P\left( {{Y \over X}} \right) = {{P(X \cap Y)} \over {P(X)}} = {2 \over 5}$$
$$P(X \cap Y) = {2 \over {15}}$$
$$P(Y) = {4 \over {15}}$$
$$P\left( {{{X'} \over Y}} \right) = {{P(Y) - P(X \cap Y)} \over {P(Y)}}$$
$$ = {{{4 \over {15}} - {2 \over {15}}} \over {{4 \over {15}}}} = {1 \over 2}$$
$$P(X \cup Y) = $$$${1 \over 3} + {4 \over {15}} - {2 \over {15}}$$
$$ = {7 \over {15}} = {7 \over {15}}$$
$$P\left( {{X \over Y}} \right) = {{P(X \cap Y)} \over {P(Y)}} = {1 \over 2}$$
$$P\left( {{Y \over X}} \right) = {{P(X \cap Y)} \over {P(X)}} = {2 \over 5}$$
$$P(X \cap Y) = {2 \over {15}}$$
$$P(Y) = {4 \over {15}}$$
$$P\left( {{{X'} \over Y}} \right) = {{P(Y) - P(X \cap Y)} \over {P(Y)}}$$
$$ = {{{4 \over {15}} - {2 \over {15}}} \over {{4 \over {15}}}} = {1 \over 2}$$
$$P(X \cup Y) = $$$${1 \over 3} + {4 \over {15}} - {2 \over {15}}$$
$$ = {7 \over {15}} = {7 \over {15}}$$
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