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Chapter 3: Q. 3.25 (page 109)

Prove directly that,

P(E∣F)=P(E∣FG)P(G∣F)+PE∣FGcPGc∣F.

Short Answer

Expert verified

By applying the definition of conditional probability,the direction starting from the right side.

Step by step solution

01

Step: 1 Conditional probability:

The potential of an event or outcome occurring dependent on the existence of a preceding event or outcome is known as conditional probability. It's computed by multiplying the likelihood of the previous occurrence by the probability of the next, or conditional, event.

02

Step: 2 Proving equation:

By conditional probability,

P(E∣FG)P(G∣F)+PE∣FGcPGc∣F=P(EFG)P(GF)×P(GF)P(F)+PEFGcPGcF×PGcFP(F)P(E∣FG)P(G∣F)+PE∣FGcPGc∣F=P(EFG)P(F)+PEFGcP(F)P(E∣FG)P(G∣F)+PE∣FGcPGc∣F=P(EF)P(F)P(E∣FG)P(G∣F)+PE∣FGcPGc∣F=P(E∣F).

03

Step: 3 Equating probability:

In total probability,

PF=P(⋅∣F)PF(E∣G)=PF(EG)PF(G)PF(E∣G)=P(EG[F)P(G∣F)PF(E∣G)=P(EFG)P(F)P(FG)P(F)PF(E∣G)=P(E∣FG).

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Most popular questions from this chapter

Suppose that an ordinary deck of 52cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the role="math" localid="1647784076635" 20thcard to appear, what is the conditional probability that the card following it is the

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