S1 - probability

2. A car dealer offers purchasers a three year warranty on a new car. He sells two models, the Zippy and the Nifty. For the first 50 cars sold of each model the number of claims under the warranty is shown in the table below.

Claim No claim
Zippy 35 15
Nifty 40 10

One of the purchasers is chosen at random. Let A be the event that no claim is made by the purchaser under the warranty and B the event that the car purchased is a Nifty.

(a) Find P(A n B).

(b) Find P(A').

Given that the purchaser chosen does not make a claim under the warranty,

(c) find the probability that the car purchased is a Zippy.

(d) Show that making a claim is not independent of the make of the car purchased.

Comment on this result.

________________________

Above is a question from the S1 January 2003 paper. I don't quite understand it fully. To me it seems that they are independent events but when part (d) said "show that making a claim is not independent of the make of the car purchased". Confused. Could anyone please help me out on this question :frown:

Thanks!

Reply 1

The Colonel

13

a. P(A n B) = 10/100 = 1/10 = 0.1
b. P(A') = 75/100 = 0.75
c. P(B'/A) = P(B'nA)/P(A) = 15/100/25/100 = 15/25 = 3/5 = 0.6
d. P(A'nB) = 0.4, P(A')P(B) = 0.75 x 0.5 = 0.375
Since P(A'nB) is not equal to P(A')p(B) --> Not independent
Therefore, one of the models is less reliable

Hope this helps!

Reply 2

azukibean

OP

6

The Colonel

a. P(A n B) = 10/100 = 1/10 = 0.1
b. P(A') = 75/100 = 0.75
c. P(B'/A) = P(B'nA)/P(A) = 15/100/25/100 = 15/25 = 3/5 = 0.6
d. P(A'nB) = 0.4, P(A')P(B) = 0.75 x 0.5 = 0.375
Since P(A'nB) is not equal to P(A')p(B) --> Not independent
Therefore, one of the models is less reliable

Hope this helps!

Hey thanks! I get it now =) the wording of the question just confused me a bit :biggrin: