Let X and Y be two events such that P(X) = (1)/(3) , P(XY) = (1)/(2) and P(Y | X) = (2)/(5) . Then

Correct answer: A. P(Y) = (4)/(15) Using formula P(X | Y) = (P( X Y ))/(P( Y )) and P(Y | X) = (P( Y X ))/(P( X )) P(X) = (1)/(3); (P(X Y))/(P(Y)) = (1)/(2); (P(Y X))/(P(X)) = (2)/(5) From this information, we get P(X Y) = (2)/(15); P(Y) = (4)/(15) Using formula P ( X Y ) = P ( X ) + P ( Y ) - P ( X Y ) ∴ P(X Y) = (1)/(3) + (4)/(15) - (2)/(15) = (7)/(15) P( ¯ X | Y) = P( - X Y) P(Y) = (P(Y) - P(X Y))/(P(Y)) ⇒ P( ¯ X | Y) = 1 - ((2)/(15))/((4)/(15)) = (1)/(2)

JEE Advanced 2017Paper 1

Exam: JEE Advanced
Year: 2017
Paper: Paper 1
Subject: Mathematics
Chapter: Probability

JEE Advanced 2017 · Paper 1 · Mathematics · Probability

Let X and Y be two events such that P(X) = (1)/(3) , P(XY) =...

Let

X

and

Y

be two events such that

P\left(X\right) = \dfrac{1}{3}

P\left(XY\right) = \dfrac{1}{2}

and

P\left(Y | X\right) = \dfrac{2}{5} .

Then

Options

A $P\left(Y\right) = \dfrac{4}{15}$
B $P\left(\overset{¯}{X} | Y\right) = \dfrac{1}{2}$
C $P\left(X \cap Y\right) = \dfrac{1}{5}$
D $P\left(X \cup Y\right) = \dfrac{2}{5}$

Correct answer

A. $P\left(Y\right) = \dfrac{4}{15}$

Step-by-step solution

Using formula

P\left(X | Y\right) = \dfrac{P( X \cap Y )}{P( Y )}

and

P\left(Y | X\right) = \dfrac{P( Y \cap X )}{P( X )}

P\left(X\right) = \dfrac{1}{3}; \dfrac{P\left(X \cap Y\right)}{P\left(Y\right)} = \dfrac{1}{2}; \dfrac{P\left(Y \cap X\right)}{P\left(X\right)} = \dfrac{2}{5}

From this information, we get

P\left(X \cap Y\right) = \dfrac{2}{15}; P\left(Y\right) = \dfrac{4}{15}

Using formula

P ( X \cup Y ) = P ( X ) + P ( Y ) - P ( X \cap Y )

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

P\left(\overset{¯}{X} | Y\right) = \dfrac{P\left(\overset{-}{X} \cap Y\right)}{P\left(Y\right)} = \dfrac{P\left(Y\right) - P\left(X \cap Y\right)}{P\left(Y\right)}

⇒ P\left(\overset{¯}{X} | Y\right) = 1 - \dfrac{\dfrac{2}{15}}{\dfrac{4}{15}} = \dfrac{1}{2}

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