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Mathematics Question on Conditional Probability

Let E1E_1 and E2E_2 be two events such that the conditional probabilities P(E1∣E2)=12P(E_1|E_2)=\frac 12, P(E2∣E1)=34P(E_2|E_1)=\frac 34 and P(E1∩E2)=18P(E_1∩E_2)=\frac 18⋅ Then:

A

P(E1∩E2)=P(E1)⋅P(E2)P(E_1∩E_2)=P(E_1)⋅P(E_2)

B

P(E1′∩E2′)=P(E1′)⋅P(E2′)P(E'_1∩E'_2)=P(E'_1)⋅P(E'_2)

C

P(E1∩E2′)=P(E1)⋅P(E2)P(E_1∩E'_2)=P(E_1)⋅P(E_2)

D

P(E1′∩E2)=P(E1)⋅P(E2)P(E'_1∩E_2)=P(E_1)⋅P(E_2)

Answer

P(E1∩E2′)=P(E1)⋅P(E2)P(E_1∩E'_2)=P(E_1)⋅P(E_2)

Explanation

Solution

P(E1E2)=12P(\frac {E_1}{E_2})=\frac 12

⇒P(E1∩E2)P(E2)=12⇒\frac {P(E_1∩E_2)}{P(E_2)}=\frac 12

P(E2E1)=34P(\frac {E_2}{E_1})=\frac 34

⇒P(E2∩E1)P(E1)=34\frac {P(E_2∩E_1)}{P(E_1)}=\frac 34

P(E1∩E2)=18P(E_1∩E_2)=\frac 18

P(E2)=14, P(E1)=16P(E_2)=\frac 14,\ P(E_1)=\frac 16

(A) P(E1∩E2)=18P(E_1∩E_2)=\frac 18 and P(E1)⋅P(E2)=124P(E_1)⋅P(E_2)=\frac {1}{24}
⇒P(E1∩E2)≠P(E1).P(E2)⇒ P(E_1∩E_2)≠P(E_1).P(E_2)

(B) P(E1′∩E2′)=1−P(E1∪E2)P(E_1'∩E_2')=1−P(E_1∪E_2)

=1−[14+16−18]=1724=1−[\frac 14+\frac 16−\frac 18]=\frac {17}{24}

P(E1′)=34⇒P(E1′)P(E2)=324P(E_1')=\frac 34⇒P(E_1')P(E_2)=\frac {3}{24}
⇒P(E1′∩E2′)≠P(E1′)⋅P(E−2)⇒ P(E_1'∩E_2')≠P(E_1')⋅P(E-2)

(C) P(E1∩E2′)=P(E1)−P(E1∩E2)P(E_1∩E_2')=P(E_1)−P(E_1∩E_2)
=16−18=124=\frac 16−\frac 18=\frac {1}{24}
P(E1)⋅P(E2)=124P(E_1)⋅P(E_2)=\frac {1}{24}
⇒P(E1∩E2′)=P(E1)⋅P(E2)⇒ P(E_1∩E_2')=P(E_1)⋅P(E_2)

(D) P(E1′∩E2)=P(E2)−P(E1∩E2)P(E_1'∩E_2)=P(E_2)−P(E_1∩E_2)
=14−18=18=\frac 14−\frac 18=\frac 18
P(E1)P(E2)=124P(E_1)P(E_2)=\frac {1}{24}
⇒P(E1′∩E2)≠P(E1).P(E2)⇒ P(E_1'∩E_2)≠P(E_1).P(E_2)

So, the correct option is (C): P(E1∩E2′)=P(E1)⋅P(E2)P(E_1∩E_2')=P(E_1)⋅P(E_2)