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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Probability

BITSAT 2023 · Paper 2 · Mathematics · Probability

If E and F are events such that 0 < P(F) < 1 , then

If E\mathrm{E} and F\mathrm{F} are events such that 0<P(F)<10 < \mathrm{P}(\mathrm{F}) < 1, then

Options

  1. AP(E∣F)+P(Eˉ∣F)=1P(E \mid F)+P(\bar{E} \mid F)=1
  2. BP(E∣F)+P(E∣Fˉ)=1P(E \mid F)+P(E \mid \bar{F})=1
  3. CP(Eˉ∣F)+P(E∣Fˉ)=1P(\bar{E} \mid F)+P(E \mid \bar{F})=1
  4. DP(E∣Fˉ)+P(Eˉ∣Fˉ)=0P(E \mid \bar{F})+P(\bar{E} \mid \bar{F})=0

Correct answer

A. P(E∣F)+P(Eˉ∣F)=1P(E \mid F)+P(\bar{E} \mid F)=1

Step-by-step solution

P(E∣F)+P(Eˉ∣F)P(E \mid F)+P(\bar{E} \mid F)
=P(E∩F)+P(Eˉ∩F)P(F)=P((E∪Eˉ)∩F)P(F)=P(F)P(F)=1=\frac{P(E \cap F)+P(\bar{E} \cap F)}{P(F)}=\frac{P((E \cup \bar{E}) \cap F)}{P(F)}=\frac{P(F)}{P(F)}=1
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