Let E₁ and E₂ be two events such that the conditional probabilities P(E₁ E₂)=1/2, P(E₂ E₁)=3/4 and P(E₁ E₂)=1/8. Then : [25-Jun-2022-Shift-1]

Correct Answer is : P(E1∩E2′)=P(E1)⋅P(E2)P(E₁ E₂')=P(E₁) · P(E₂)P(E1∩E2′)=P(E1)⋅P(E2)

P(E₁ E₂)=P(E₁) · P(E₂)

P(E₁' E₂')=P(E₁') · P(E₂)

P(E₁ E₂')=P(E₁) · P(E₂)

P(E₁' E₂)=P(E₁) · P(E₂)

Correct Answer is : P(E1∩E2′)=P(E1)⋅P(E2)P(E₁ E₂')=P(E₁) · P(E₂)P(E1∩E2′)=P(E1)⋅P(E2)

Test Index

JEE Main 25 June 2022 Shift 1 Paper

Section: Mathematics

Question : 10 of 90

Marks: +1, -0

Let

E_1

E_2

P(E_1 \mid E_2)=\frac{1}{2}, P(E_2 \mid E_1)=\frac{3}{4}

P(E_1 \cap E_2)=\frac{1}{8}

Solution: 👈: Video Solution

P\left(\frac{E_1}{E_2}\right)=\frac{1}{2} \Rightarrow \frac{P(E_1 \cap E_2)}{P(E_2)}=\frac{1}{2}

P\left(\frac{E_2}{E_1}\right)=\frac{3}{4} \Rightarrow \frac{P(E_2 \cap E_1)}{P(E_1)}=\frac{3}{4}

P(E_1 \cap E_2)=\frac{1}{8}

P(E_2)=\frac{1}{4}, P(E_1)=\frac{1}{6}

(A)

P(E_1 \cap E_2)=\frac{1}{8}

and

P(E_1) \cdot P(E_2)=\frac{1}{24}

\Rightarrow P(E_1 \cap E_2) \neq P(E_1) \cdot P(E_2)

(B)

P(E_1' \cap E_2')=1-P(E_1 \cup E_2)

=1-\left[\frac{1}{4}+\frac{1}{6}-\frac{1}{8}\right]=\frac{17}{24}

P(E_1')=\frac{3}{4} \Rightarrow P(E_1') P(E_2)=\frac{3}{24}

\Rightarrow P(E_1' \cap E_2') \neq P(E_1') \cdot P(E_2)

(C)

P(E_1 \cap E_2')=P(E_1)-P(E_1 \cap E_2)

=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}

P(E_1) \cdot P(E_2)=\frac{1}{24}

\Rightarrow P(E_1 \cap E_2')=P(E_1) \cdot P(E_2)

(D)

P(E_1' \cap E_2)=P(E_2)-P(E_1 \cap E_2)

=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}

P(E_1) P(E_2)=\frac{1}{24}

\Rightarrow P(E_1' \cap E_2) \neq P(E_1) \cdot P(E_2)

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