Conditional probability question

Question

An insurance company classies drivers as class X, Y or Z. Experience indicates that

the probability a class X driver has at least one accident in any one year is 0.01, while

the corresponding probabilities for Y and Z are 0.05 and 0.1 respectively. The class

of a new driver who applies to them for cover is not known, but from experience they

have found that 55% of applicants are class X, 40% are class Y , and 5% are class Z.

(a) What is the probability of a new client having an accident in their rst year?

(b) If a new client has no accidents in their rst year, what is the probability that

they are a class Z driver?

(c) What is the probability of a new client having no accidents in their rst 4 years?

(You may assume that conditional on the class of driver, the occurrence of acci-

dents is independent between years.)

Answer 1

The table method is useful with two sets of mutually exclusive and collectively exhaustive events, such as this problem and many others. One set is the possible class = {X, Y, Z} which are the rows of the table. 

The other set is {A, A'} for Accident and No Accident, for the columns. We are given the marginal probabilities for the classes, which are the totals for the rows.

Calculate the joint probability P(X and A) = P(A | X)*P(X) = 0.01*.55 = .0055, which is the table entry for row X and column A. For row X and column A' the entry is 0.55 - 0.0055 = 0.5445. Similarly for rows Y and Z. The completed table is shown below.

 

(a): Total probability of an accident is the sum for column A = .0305

(b): Conditional probability P(Z | A') = P(A AND A') / P(A') = .045/.9695 = 0.046416

(c): For class X, the probability of no accident for 4 years (given years are conditionally independent) is

         ( P(A' | X) )^4 = .087901, and similarly for the other two classes. 

The joint probability P(no accident for 4 years AND X) = .087901 * .55 = .04835. 

Adding the three joint probabilities gives the answer, 0.056686

 

 

 

C\A......A=accident......A'...........Total......P(A'^4|C)......Joint
X..........0.0055...........0.5445.....0.55.......0.087901.......0.04835
Y..........0.02...............0.38.........0.4........0.020851.......0.00834
Z..........0.005.............0.045........0.05......4.1E-06.........2.1E-07
Total.....0.0305...........0.9695.......1..............................0.05669
..............................
Given:..................Answers............
0.55......P(X)............(a)=P(A)=...............0.0305
0.40......P(Y)............(b)=P(Z|A')=............0.046416
0.05......P(Z)............(c)=P(A' A' A' A')=....0.056686
0.01......P(A|X)
0.05......P(A|Y)
0.10......P(A|Z)