6-BAYES’S FORMULA

Let E and F be events so

P(E|F)=

We may express E as𝐸 =(𝐸∩𝐹)  ∪  (𝐸∩ )

As 𝐸 ∩ 𝐹 and 𝐸 ∩ are clearly mutually exclusive, we have, by Axiom 3, 

This equation  states that the probability of the event E is a weighted

average of the conditional probability of E given that F has

occurred and the conditional probability of E given that F has not occurred—each

conditional probability being given as much weight as the event on which 

it is conditioned has of occurring. This is an extremely useful formula,

because its use often enables us to determine the probability of an event by first

“conditioning” upon whether or not some second event has occurred.

That is, there are many instances in which it is difficult  to  compute 

the  probability  of  an  event  directly,  but  it  is straightforward  to

compute  it  once  we  know  whether  or  not  some second event has occurred.

An insurance company believes that people can be divided into two classes: 

those  who  are  accident  prone  and  those  who  are  not. 

The company’s statistics show that an accident-prone person will have an

accident at sometime within a fixed 1-year period with probability .

4, whereas this probability decreases to .2 for a person who is not accident prone. 

If we assume that 30 percent of the population is accident prone, what is the probability

that a new policyholder will have an accident within a year of purchasing a policy? 

We shall obtain the desired probability by first conditioning upon whether

or not the policyholder is accident prone.

Let denote the event that the policyholder will have an accident within a year 

of purchasing the policy, and let A denote the event that the policyholder is accident prone.

Hence, the desired probability is given by

Suppose  that  a  new  policyholder  has  an  accident  within 

a  year  of purchasing a policy.

What is the probability that he or she is accident prone? Solution. The desired probability is

Suppose we are interested in diagnosing cancer in patients who visit a chest clinic.

Let C represent the event "Person has cancer" Let S represent the event

"Person is a smoker"

*We know the probability of the prior event P(C) = 0.1 on

the basis of past data (10% of the patients entering the clinic turn out to have cancer).

*We know P(S|C) by checking from our record the proportion of 

smokers among those diagnosed with cancer. Suppose P(S|C) = 0.8

*We know P(𝑆|)by

checking from our record the proportion of smokers  among  those  not  diagnosed  with  cancer.

Suppose P(𝑆|) = 0.466 P(S) = (0.8)(0.1) + (0.466)(0.9) = 0.5 P(C|S) =  =0.16 Thus,

in light of the evidence that the person is a smoker we revise our prior

probability from 0.1 to a posterior probability of 0.16. This is a significance increase,

but it is still unlikely that the person has cancer. 

You  ask  your  neighbor  to  water  a  sickly  plane  while  you  are  on vacation.

Without water it will die with probability 0.8; with water it 

will die with probability 0.15. You are 90 percent certain that your

neighbor will remember to water the plant. 

(a)What is the probability that the plant will be alive when you return?

(b)If it is dead, what is the probability your neighbor forgot to water it? 

According to the Arizona Chapter of the American Lung Association,

7.0% of the population has lung disease. 

Of those having lung disease, 90% are smokers; of those not having lung disease,

25.3% are smokers. Determine the probability that a randomly selected smoker has lung disease.

Solution:

A certain federal agency employs three consulting firms (A, B, and C) with  probabilities 

0.40,  0.35,  and  0.25,  respectively.  From  past experience it is known

that the probability of cost overruns for the firms are  0.05,  0.03,  and  0.15  respectively.

Suppose  a  cost  overrun  is experienced by the agency.

(a) What is the probability that the consulting firm involved is company C?

(b) What is the probability that it is company A?