Question

Let X and Y be two events such that P(X|Y) = $\dfrac{1}{2}$, P(Y|X) = $\dfrac{1}{3}$, ${\text{P(X}} \cap {\text{Y)}}$= $\dfrac{1}{6}$. Which of the following is correct?
A. P(XUY) = $\dfrac{2}{3}$
B. X and Y are independent.
C. X and Y are not independent.
D. ${\text{P}}\left( {{{\text{X}}^{\text{C}}} \cap {\text{Y}}} \right){\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}$.

Answer 1

Hint: In order to solve this problem you need to know about the term P(X|Y) and P(Y|X). Then you have to evaluate P(XUY) to check it. As we know that P(X|Y) = $\dfrac{{{\text{P(X}} \cap {\text{Y)}}}}{{{\text{P(Y)}}}}$ and P(Y|X) = $\dfrac{{{\text{P(Y}} \cap {\text{X)}}}}{{{\text{P(X)}}}}$. Doing this and finding the value of P(X) and P(Y) will solve your problem.

Complete step-by-step answer:
The given things in the question are P(X|Y) = $\dfrac{1}{2}$, P(Y|X) = $\dfrac{1}{3}$, ${\text{P(X}} \cap {\text{Y)}}$= $\dfrac{1}{6}$.
So, we solve P(X|Y) = $\dfrac{{{\text{P(X}} \cap {\text{Y)}}}}{{{\text{P(Y)}}}}$
On putting the values and solving we get,
P(Y) = $\dfrac{{\dfrac{1}{6}}}{{\dfrac{1}{2}}} = \dfrac{1}{3}$
Similarly we will solve P(Y|X) = $\dfrac{{{\text{P(Y}} \cap {\text{X)}}}}{{{\text{P(X)}}}}$.
On putting the values and solving we get,
P(X) = $\dfrac{{\dfrac{1}{6}}}{{\dfrac{1}{3}}} = \dfrac{1}{2}$
We know that ${\text{P(X}} \cup {\text{Y) = P(X) + P(Y) - P(X}} \cap {\text{Y)}}$
On putting the values we get,

$
  {\text{P(X}} \cup {\text{Y)}} = \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{6} \\
  {\text{P(X}} \cup {\text{Y)}} = \dfrac{5}{6} - \dfrac{1}{6} = \dfrac{4}{6} = \dfrac{2}{3} \\
$
So, Option A is correct.
We know that if P(X).P(Y) = ${\text{P(X}} \cap {\text{Y)}}$ then the events are independent.
We can observe from the above data that the equation P(X).P(Y) = ${\text{P(X}} \cap {\text{Y)}}$ is satisfied.
That is ${\text{P(X}} \cap {\text{Y) = }}\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ and it is also given ${\text{P(X}} \cap {\text{Y) = }}\dfrac{1}{6}$.
Therefore option B is correct.
We will not check Option C since the events are independent.
IF we consider option D it is said that ${\text{P}}\left( {{{\text{X}}^{\text{C}}} \cap {\text{Y}}} \right){\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}$
As we know that ${{\text{X}}^{\text{C}}} \cap \,{\text{Y = Y}} - {\text{X}}$ that is the event of Y excluding that of X.
But we know that X and Y are independent so $P\left( {{{\text{X}}^{\text{C}}} \cap \,{\text{Y}}} \right) = P\left( {{X^C}} \right).P\left( Y \right) = \dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ since $P\left( {{X^C}} \right) = 1 - P\left( X \right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$.
Therefore from the above data we can clearly see that options A and B are correct.

So, the correct answer is “Option A and B”.

Note: When you get to solve such problems you need to know the general formulas of the probabilities and the terms denoted like P(Y|X) and P(X|Y) and you should also know the condition of independent events and the formula ${\text{P(X}} \cup {\text{Y) = P(X) + P(Y) - P(X}} \cap {\text{Y)}}$. Events A and B are independent if the equation ${\text{P(X}} \cap {\text{Y)}}$ = P(A).P(B) holds true. You can use the equation to check if events are independent; multiply the probabilities of the two events together to see if they equal the probability of them both happening together.