2.3 Conditional Probability

Example 2.3.1 An experiment consists of dealing two cards from a pack, one after theother. Find the probability that both cards are diamonds.

Solution : Events :

1D : 1st card is a diamond 2D : 2nd card is a diamond

We need P (ID 2D).We have P (1D) = 1/4. This means - we expect that the 1st card will be a diamond in 1/4of all repetitions of the experiment. In how many of these cases do we expect that the 2ndcard will be a diamond?

After the first is dealt there are 51 cards remaining, of which 12 are diamonds. Thus,in 12/51 of those 1/4 of all cases in which the 1st card is a diamond, the second is also adiamond, i.e. in

12

51 1

4=

1

17

of all cases.We conclude that P (1D and 2D) = 117 .

Terminology: 12/51 = 4/17 is the conditional probability of 2D given the 1D occurs. It isdenoted by P (2D|1D).

This is not the same as P (2D) which is 1/4.

Nor is it the same as P (2D|1D) which is 13/51.

Formula 2.3.2 Let A and B be events associated to an experiment. Then

P (A and B) = P (A B) = P (A)P (B|A) = P (B)P (A|B).

Alternatively

P (B|A) = P (A B)P (A)

andP (A|B) = P (A B)P (B)

.

Example 2.3.3 Two fair dice are rolled. Find the probability that the total score is at least8, given that it is an even number.

Solution: Events : A - total score 8; B - total score even. We need P (A|B) and we know

P (A|B) = P (A B)P (B)

.

There are 36 outcomes in 18 of which the total score is even. Thus P (B) = 18/36 = 1/2.There are 9 outcomes in which the total score is even and at least 8. Thus

P (A B) = 9/36 = 1/4.

Now

P (A|B) = 1/41/2

= 1/2.

Note: This is not equal to P (A) which is 15/36.Exercise: Show that P (B|A) = 3/5.

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Definition 2.3.4 Events A and B are called independent if

P (A B) = P (A)P (B).

This means P (B) = P (B|A) and P (A) = P (A|B), i.e. the probability of one of these eventsoccurring is unaffected by whether the other occurs or not.

Example 2.3.5 A factory has machines A and B making 60% and 40% respectively oftotal production. 3% of the items produced by Machine A and 5% of the items produced byMachine B are defective.

(a) Find the probability that a randomly selected item is defective.

(b) Given that an item is defective, find the probability that it came from Machine A.

Solution: Events :

A - (randomly selected) item came from Machine AB - (randomly selected) item came from Machine BD - (randomly selected) item is defective

We know : P (A) = 0.6, P (B) = 0.4, P (D|A) = 0.03, P (D|B) = 0.05.(a) We need P (D). The event D is the union of the mutually exclusive events D A

(defective and from A) and D B (defective and from B). Thus

P (D) = P (D A) + P (D B)= P (A)P (D|A) + P (B)P (D|B)= 0.6(0.03) + 0.4(0.05)

= 0.038

(Alternatively - 3% of the 60% of items from Machine A are defective, or 1.8% of allitems, and 5% of items from Machine B, or 2% of all items, are defective - this is 3.8%of all items in total.)

(b) We need P (A|D). This is found by

P (A|D) = P (A D)P (D)

=P (A)P (D|A)

P (A)P (D|A) + P (B)P (D|B)=

0.6(0.03)

0.038= 0.4737

Example 2.3.6 A motor insurance company insures drivers in age groups A, B and C.40% of the companys customers are in age group A, 25% are in B and 35% are in C.

The companys records show that on average 2% of customers in age group A, 1% ofthose in age group B, and 1.5% of those in age group C make a claim each year.

(a) What is the probability that a randomly selected driver will make a claim next year?

(b) If a driver is selected at random, what is the probability that this person is from agegroup C and will not make a claim in the next year?

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(c) Given that a driver has made a claim in the past year, what is the probability that thisperson is from age group C?

Solution: Events :

A : Randomly selected person is from age group A.

B : Randomly selected person is from age group B.

C : Randomly selected person is from age group C.

Cl : Randomly selected person will make a claim in the next year.

We knowP (A) = 0.4 P (B) = 0.25 P (C) = 0.35P (Cl|A) = 0.02 P (Cl|B) = 0.01 P (Cl|C) = 0.015

(a) P (Cl) = P (Cl A) + P (Cl B) + P (Cl C)= P (A)P (Cl|A) + P (B)P (Cl|B) + P (C)P (Cl|C)= 0.04(0.02) + 0.25(0.01) + 0.35(0.015)= 0.01575

(b) P (Cl C) = P (C)P (Cl|C)= 0.35(0.985) = 0.34475

(c) P (A|Cl) = P (A Cl)P (Cl)

=P (A)P (Cl|A)

P (Cl)

0.4(0.02)

0.01575= 0.5079

In the above examples we have used Bayess Theorem, which states the following.

Theorem 2.3.7 (Bayes) Let E1, . . . , Ek be subsets of a sample space S with the propertythat every element of S belongs to exactly one of the Ei. Let A be any event (subset of S).Then

P (Ei|A) = P (Ei)P (A|Ei)kj=1 P (Ej)P (A|Ej)

.

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