Let E _ 1 and E _ 2 be two events such that the conditional probabilities P( E _ 1 ∣ E _ 2 ) = (1)/(2) , P( E _ 2 ∣ E _ 1 ) = (3)/(4) and P( E _ 1 E _ 2 ) = (1)/(8) . Then

Correct answer: C. P( E _ 1 E _ 2 ^ ' ) = P( E _ 1 ) · P( E _ 2 ) P( E _ 1 E _ 2 ) P( E _ 2 ) = (1)/(2), P( E _ 2 E _ 1 ) P( E _ 1 ) = (3)/(4), P( E _ 1 E _ 2 ) = (1)/(8) ⇒ P( E _ 2 ) = (1)/(4), P( E _ 1 ) = (1)/(6), P( E _ 1 E _ 2 ) = (1)/(6) + (1)/(4) - (1)/(8) = (7)/(24) P( E ^ ' E _ 2 ^ ' ) = P ( E _ 1 E _ 2 ) ^ ' = 1 - (7)/(24) = (17)/(24) P( E _ 2 ^ ' ) = (3)/(4) ⇒ P( E _ 1 E _ 2 ) = P( E _ 1 )P( E _ 2 ^ ' ) P( E _ 1 E _ 2 ^ ' ) = P( E _ 1 ) - P( E _ 1 E _ 2 ) = (1)/(6) - (1)/(8) = (1)/(24) P( E _ 1 E _ 2 ^ ' ) = P( E _ 1 ) · P( E _ 2 )

JEE Main 202225 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 25 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Probability

JEE Main 2022 · 25 June · Morning Shift · Mathematics · Probability

Let E _ 1 and E _ 2 be two events such that the conditional...

Let

{E}_{1}

and

{E}_{2}

be two events such that the conditional probabilities

P\left({E}_{1} ∣ {E}_{2}\right) = \dfrac{1}{2}

P\left({E}_{2} ∣ {E}_{1}\right) = \dfrac{3}{4}

and

P\left({E}_{1} \cap {E}_{2}\right) = \dfrac{1}{8}

. Then

Options

A $P\left({E}_{1} \cap {E}_{2}\right) = P\left({E}_{1}\right) \cdot P\left({E}_{2}\right)$
B $P\left({E}_{1}^{'} \cap {E}_{2}^{'}\right) = P\left({E}_{1}^{'}\right) \cdot P\left({E}_{2}\right)$
C $P\left({E}_{1} \cap {E}_{2}^{'}\right) = P\left({E}_{1}\right) \cdot P\left({E}_{2}\right)$
D $P\left({E}_{1} \cup {E}_{2}\right) = P\left({E}_{1}\right)P\left({E}_{2}\right)$

Correct answer

C. $P\left({E}_{1} \cap {E}_{2}^{'}\right) = P\left({E}_{1}\right) \cdot P\left({E}_{2}\right)$

Step-by-step solution

$\dfrac{P\left({E}_{1} \cap {E}_{2}\right)}{P\left({E}_{2}\right)} = \dfrac{1}{2}, \dfrac{P\left({E}_{2} \cap {E}_{1}\right)}{P\left({E}_{1}\right)} = \dfrac{3}{4}, P\left({E}_{1} \cap {E}_{2}\right) = \dfrac{1}{8}$

$⇒ P\left({E}_{2}\right) = \dfrac{1}{4}, P\left({E}_{1}\right) = \dfrac{1}{6}, P\left({E}_{1} \cup {E}_{2}\right) = \dfrac{1}{6} + \dfrac{1}{4} - \dfrac{1}{8} = \dfrac{7}{24}$

$P\left({E}^{'} \cap {E}_{2}^{'}\right) = P{\left({E}_{1} \cup {E}_{2}\right)}^{'} = 1 - \dfrac{7}{24} = \dfrac{17}{24}$

$P\left({E}_{2}^{'}\right) = \dfrac{3}{4} ⇒ P\left({E}_{1} \cap {E}_{2}\right) = P\left({E}_{1}\right)P\left({E}_{2}^{'}\right)$

$P\left({E}_{1} \cap {E}_{2}^{'}\right) = P\left({E}_{1}\right) - P\left({E}_{1} \cap {E}_{2}\right) = \dfrac{1}{6} - \dfrac{1}{8} = \dfrac{1}{24}$

$P\left({E}_{1} \cap {E}_{2}^{'}\right) = P\left({E}_{1}\right) \cdot P\left({E}_{2}\right)$

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