Let E1 and E2 be two events such that the conditional probabilities P ( E 1 mid E 2)=(1/2), P ( E 2 mid E 1)=(3/4) and P ( E 1 ∩ E 2)=(1/8). Then:

Question

Let E1 and E2 be two events such that the conditional probabilities P ( E 1 mid E 2)=(1/2), P ( E 2 mid E 1)=(3/4) and P ( E 1 ∩ E 2)=(1/8). Then: (A) P ( E 1 ∩ E 2)= P ( E 1) ⋅ P ( E 2) (B) P ( E 1 prime ∩ E 2 prime)= P ( E 1 prime) ⋅ P ( E 2) (C) P ( E 1 ∩ E 2 prime)= P ( E 1) ⋅ P ( E 2) (D) P ( E 1 prime ∩ E 2)= P ( E 1) ⋅ P ( E 2)

Tardigrade Admin · Accepted Answer

(A) P ( E 1) ⋅ P ( E 2)=(1/6) ⋅ (1/4)=(1/24) ≠ P ( E 1 ∩ E 2) (B) P ( E 1 prime ∩ E 2 prime) =1- P ( E 1 ∪ E 2) .=1-( P ( E 1)+ P ( E 2)- P 1 E 1 ∩ E 2)) =1-((1/6)+(1/4)-(1/8))=(17/24) P ( E 1 prime) P ( E 2)=(5/6) × (1/4)=(5/24) (C) P ( E 1 ∩ E 2 prime)= P ( E 1)- P ( E 1 ∩ E 2) =(1/6)-(1/8)=(1/24) (D) P ( E 1 prime ∩ E 2)= P ( E 2)- P ( E 1 ∩ E 2) =(1/4)-(1/8)=(1/8)

Q. 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:

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

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

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

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

Q. Let E1​ and E2​ be two events such that the conditional probabilities P(E1​∣E2​)=21​, P(E2​∣E1​)=43​ and P(E1​∩E2​)=81​. Then:

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​)

Q. 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:

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

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

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

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