Heather,

We need a matrix where the items are each Probabilities (P), P(i,j), where P(1,2) is the probability of going from group 2 to group 1 in one step so P(F,C) is the probability from shifting from a Chevy to a Ford and P(F,F) is the probability of staying with a ford. So your matrix needs to be

P(C,C) P(C,F) P(C,H)

P(F,C) P(F,F) P(F,H)

P(H,C) P(H,F) P(H,H)

We're given the probabilities of switching, so the remaining probability has to be the probability of staying with the same kind of car, and the remaining probability must make the total probability of switching or staying add up to one, so we can calculate the underlined values.

Current Car

N C F H

E C C .45 .35 .20

X A F .25 .50 .45

T R H .30 .15 .35

For your second problem, there are two approaches.

The first is to do some matrix algebra. It turns out that state x at time t+1 is equal to P (the transition matrix) times the state x at time t or

x^{(t+1)}=Px^{(t)}

and thus if t=0 is when everyone has a Ford, t=1 is their first purchase and t=2 is their second purchase, we can find x^{(2)} = Px^{(1)} = P(Px^{(0)})=P^{2}x^{(0)}

So if you can do Matrix multiplication twice, you can square P (i.e. multiply it by itself) and then multiply it by its initial condition matrix, which would be everyone owning a Ford or a three item vertical matrix like this

C 0

F 1

H 0

Explaining matrix multiplication is another topic (I used the second method and an online calculator to compute this matrix),but P-squared should be this

0.35 0.3625 0.3175

0.3725 0.405 0.4325

0.2775 0.2325 0.25

but it would produce this final vertical matrix of the probability of what their second car purchased would be.

C 0.3625

F 0.405

H 0.2325

Alternatively (and more intuitively), you can use a decision tree.

The current car is a Ford (given), so the person can stay with a Ford (.50), or switch to a Chevy (.35) or a Honda (.15) when they buy the first car, and then you need to repeat the probabilities for each one

F-->F(.50)-->F(.50)=0.2500

F-->F(.50)-->C(.35)=0.1750

F-->F(.50)-->H(.15)=0.0750 **Ended up with a Honda**

F-->C(.35)-->C(.45)=0.1575

F-->C(.35)-->F(.25)=0.0875

F-->C(.35)-->H(.30)=0.1050 **Ended up with a Honda**

F-->H(.15)-->H(.35)=0.0525 **Ended up with a Honda**

F-->H(.15)-->F(.45)=0.0675

F-->H(.15)-->C(.20)=0.0300

------------------------------

=1

So the probability of ending up with a Honda is

.075 + .105 + .0525 = 0.2325