/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 3.25 Prove directly that,P(E∣F)=P(E... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a class, there are 4 first-year boys, 6 first-year girls, and 6 sophomore boys. How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining p, the probability that each hand has an ace. Let Ei be the event that I the hand has exactly one ace. Determine p = P(E1E2E3E4) by using the multiplication rule.

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer the question, pointing out that if A knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from 1 3 to 1 2 because he would then be one of two prisoners. What do you think of the jailer’s reasoning?

A and B play a series of games. Each game is independently won by A with probability p and by B with probability 1− p. They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series.

(a) Find the probability that a total of 4games are played.

(b) Find the probability that A is the winner of the series

On the morning of September 30,1982, the won–lost records of the three leading baseball teams in the Western Division of the National League were as follows:

Each team had 3games remaining. All 3of the Giants’ games were with the Dodgers, and the 3remaining games of the Braves were against the San Diego Padres. Suppose that the outcomes of all remaining games are independent and each game is equally likely to be won by either participant. For each team, what is the probability that it will win the division title? If two teams tie for first place, they have a playoff game, which each team has an equal chance of winning.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.