Stats please help

EMPhillips1989 · 2008-02-01 06:08:32

A family that owns two cars is selected at random
Let A={the older car is a BMW} B={the newer car is a BMW}
If P(A)=0.8 P(B)=0.6 P(AnB)=0.4

does anyone know how i would go about answering the following questions??

compute:
1) the probability that the newer car is a BMW given that the older car is a BMW??
2) the probability that the older car is a BMW given that the newer car is NOT a BMW???

i think it's the phrasing of the question which im getting confused on if anyone could please help i will be very grateful xxxx

mathsyperson · 2008-02-01 07:29:46

If you're asked to find P(X) and told Y, then you need to work out [P(XnY) ÷ P(Y)].

So for the first question, P(XnY) would be the probability that both cars are BMWs, which you're told is 0.4.
P(Y) would be the probability that the old car is a BMW, and you're told that is 0.8.

Therefore, your probability is 0.4/0.8 = 0.5.

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The second question is a little trickier.
First you want the probability that the older car is a BMW and the newer car isn't, or P(AnB').
You can get this by using P(A) = P(AnB) + P(AnB').
Plug in what you know: 0.8 = 0.4 + P(AnB').
Therefore, P(AnB') = 0.4.

Now, the probability the the newer car isn't a BMW is 1-P(B) (because the newer car is definitely either a BMW or not a BMW). This works out as 1 - 0.6 = 0.4.

Therefore, the probability that the older car is a BMW, given that the newer one isn't, is 0.4/0.4 = 1.

Apparently every family with two cars has a BMW. Crazy but true.

EMPhillips1989 · 2008-02-01 08:13:28

Thankyou very much you've really helped me understand the question alot better!!!!!

JaneFairfax · 2008-02-01 08:32:33

Intuitively, you can tell that the answer to (2) is going to be 1 if you notice that

So at least one of the cars must be a BMW. Hence, if one of the cars is not a BMW, the other must be.

Now try computing the probability that the newer car is a BMW given that the old one isnt. You get

mathsyperson · 2008-02-01 11:37:22

Questions like this are helped immensely by Venn Diagrams as well.
They aren't enough to answer the question (you need to talk in terms of P(x) and so on), but they're a very useful visual aid and will help you understand what's going on in each question.

Math Is Fun Forum

#1 2008-02-01 06:08:32

Stats please help

#2 2008-02-01 07:29:46

Re: Stats please help

#3 2008-02-01 08:13:28

Re: Stats please help

#4 2008-02-01 08:32:33

Re: Stats please help

#5 2008-02-01 11:37:22

Re: Stats please help

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