Let E_1 and E_2 be two events such that the conditional probabilities PE_1∣E_2=1/2, PE_2∣E_1=3/4 and PE_1∩E_2=1/8. Then

Correct answer: (C) PE_1∩E2'=PE_1·PE_2. (PE_1∩E_2)/PE_2=1/2, (PE_2∩E_1)/PE_1=3/4, PE_1∩E_2=1/8 ⇒ PE_2=1/4, PE_1=1/6, PE_1∪E_2=1/6+1/4-1/8=7/24 PE^(')∩E2'=PE1∪E2^(')=1-7/24=17/24 PE2'=3/4 ⇒PE_1∩E_2=PE_1PE2' PE_1∩E2'=PE_1-PE_1∩E_2=1/6-1/8=1/24 PE_1∩E2'=PE_1·PE_2

Q39 / 90 · JEE Main 2022 (25 Jun Shift 1) · Mathematics

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P (E_{1} ∣ E_{2}) = \frac{1}{2}$ , $P (E_{2} ∣ E_{1}) = \frac{3}{4}$ and $P (E_{1} \cap E_{2}) = \frac{1}{8}$ . Then

A
$P (E_{1} \cap E_{2}) = P (E_{1}) \cdot P (E_{2})$
B
$P (E_{1}^{'} \cap E_{2}^{'}) = P (E_{1}^{'}) \cdot P (E_{2})$
Correct answer
C
$P (E_{1} \cap E_{2}^{'}) = P (E_{1}) \cdot P (E_{2})$
D
$P (E_{1} \cup E_{2}) = P (E_{1}) P (E_{2})$

Solution

$\frac{P (E_{1} \cap E_{2})}{P (E_{2})} = \frac{1}{2}, \frac{P (E_{2} \cap E_{1})}{P (E_{1})} = \frac{3}{4}, P (E_{1} \cap E_{2}) = \frac{1}{8}$

$\Rightarrow P (E_{2}) = \frac{1}{4}, P (E_{1}) = \frac{1}{6}, P (E_{1} \cup E_{2}) = \frac{1}{6} + \frac{1}{4} - \frac{1}{8} = \frac{7}{24}$

$P (E^{'} \cap E_{2}^{'}) = P {(E_{1} \cup E_{2})}^{'} = 1 - \frac{7}{24} = \frac{17}{24}$

$P (E_{2}^{'}) = \frac{3}{4} \Rightarrow P (E_{1} \cap E_{2}) = P (E_{1}) P (E_{2}^{'})$

$P (E_{1} \cap E_{2}^{'}) = P (E_{1}) - P (E_{1} \cap E_{2}) = \frac{1}{6} - \frac{1}{8} = \frac{1}{24}$

$P (E_{1} \cap E_{2}^{'}) = P (E_{1}) \cdot P (E_{2})$

Hence, the correct answer is option: 'C'.

Let E1 and E2 be two events such that the conditional probabilities PE1∣E2=12, PE2∣E1=34 and PE1∩E2=18. Then

Solution

Relevant Questions

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P (E_{1} ∣ E_{2}) = \frac{1}{2}$ , $P (E_{2} ∣ E_{1}) = \frac{3}{4}$ and $P (E_{1} \cap E_{2}) = \frac{1}{8}$ . Then