Question

Let X and Y be two events such that$P\left( X \right){\text{ }} = \dfrac{1}{3}$, $P\left( {X/Y} \right){\text{ }} = \dfrac{1}{2}$and$P{\text{ }}\left( {Y/X} \right){\text{ }} = \dfrac{2}{5}$.Then
(A). $P\left( {X \cup Y} \right) = \dfrac{2}{5}$
(B). $P(Y) = \dfrac{4}{{15}}$
(C). $P\left( {X'/Y} \right) = \dfrac{1}{2}$
(D). $P\left( {X \cap Y} \right) = \dfrac{1}{5}$

Answer 1

To solve the question, at first we have to apply simple probability formulae to estimate$P(Y)$,$P\left( {X \cap Y} \right)$,$P\left( {X \cup Y} \right)$,$P(Y)$, $P(X')$and$P\left( {X'/Y} \right)$respectively. Finally we will choose the correct option which matches the estimation value.

Complete step-by-step answer :
Given that the
$P\left( X \right){\text{ }} = \dfrac{1}{3}$ ……………………………… (1)
$P\left( {X/Y} \right){\text{ }} = \dfrac{1}{2}$ ……………………………… (2)
$P{\text{ }}\left( {Y/X} \right){\text{ }} = \dfrac{2}{5}$ ……………………………….. (3)
We know the Baye’s formula which is given by,

\Rightarrow P\left( {X/Y} \right) = \dfrac{{P\left( {Y/X} \right)P(X)}}{{P(Y)}} \\\ \Rightarrow P(Y) = \dfrac{{P\left( {Y/X} \right)P(X)}}{{P\left( {X/Y} \right)}} \\\ $$ ………………………………………….. (4) Substituting the values of eq. (1), (2) and (3) in eq. (4) we will get,

P(Y) = \dfrac{{P\left( {Y/X} \right)P(X)}}{{P\left( {X/Y} \right)}} \\
= \dfrac{{\dfrac{2}{5} \times \dfrac{1}{3}}}{{\dfrac{1}{2}}} \\
= \dfrac{4}{{15}} \\

We know that, $$P\left( {X \cap Y} \right) = P(Y/X) \cdot P(Y)$$ ………………… (6) Substituting the values of eq. (1), and (3) in eq. (6) we will get, $$P\left( {X \cap Y} \right) = \dfrac{2}{5} \times \dfrac{1}{3} = \dfrac{2}{{15}}$$ ………………………………. (7) Again we know that, $$P\left( {X \cup Y} \right) = P(X) + P(Y) - P\left( {X \cap Y} \right)$$ ………………………. (8) Substituting the values of eq. (1), (5) and (7) in eq. (8) we will get,

P\left( {X \cup Y} \right) = \dfrac{1}{3} + \dfrac{4}{{15}} - \dfrac{2}{{15}} \\
= \dfrac{{5 + 4 - 2}}{{15}} \\
= \dfrac{7}{{45}} \\

Now the expression for $$P\left( {X'/Y} \right)$$ is given by $$P\left( {X'/Y} \right) = \dfrac{{P\left( {X' \cap Y} \right)}}{{P(Y)}} = \dfrac{{P(Y) - P(X \cap Y)}}{{P(Y)}}$$ ………………………………………. (10) Substituting the values of eq. (5) and (7) in eq. (12), we will get, $$P\left( {X'/Y} \right) = \dfrac{{\dfrac{4}{{15}} - \dfrac{2}{{15}}}}{{\dfrac{4}{{15}}}} = \dfrac{1}{2}$$ ……………………………………… (13) Here we get among the options given only options (B) and (C) match with the estimated values. **Hence options (B) and (C) are correct.** **Note** : For determining the conditional probability Baye’s formula is useful. It states that probability of an event A is based on the sum of the conditional probabilities of event A given that event B has or has not occurred and the most important is that the events A and B must be independent, mathematically $$p(B) = p\left( {B\left| A \right.} \right)p(A) + p\left( {B\left| {{A^c}} \right.} \right)p({A^c})$$