On Mon, 10 Jan 2000 [EMAIL PROTECTED] wrote:
> I was wondering if someone would be able to help me out with my
> Introduction to Statistics course I am taking. I am having problems
> with the following question :
>
> A survey of 300 senior high school students from St. Albert found that
> 180 of them either owned a car or a CD player. A further breakdown of
> the data showed that 30% of the students owned a CD player but not a
> car, and 20% of the students owned a car but not a CD player.
>
> If a senior high student is picked at random, from this group, what is
> the probability that the student owned both a car and a CD player?
>
> The answer they give is ( and I have no idea why ) :
At this point you should focus on setting out the information you have,
preferabvly in a way sufficiently systematic that you can make some sense
out of it. Only two possessions are mentioned: car and CD player.
There are 300 persons all together; 180 own either a car or a CD player;
30% of the 300 (= 90 persons) own a CD player and do NOT own a car;
20% of the 300 (= 60 persons) own a car and do NOT own a CD player.
This accounts for 150 persons; there must therefore be 30 persons
(180 - 150 = 30; I trust that part's clear enough?) who own BOTH a car
AND a CD player.
You can set this out more clearly in a table, which can contain
either numbers of persons or percentages of the total (but not a mixture
of counts and percentages!); this is how far we've got, using counts.
Own a Don't own
car a car TOTAL
Own a CD player: 30 90
Don't own a CD player: 60
TOTAL 300
All the other values in the table can be deduced by addition and/or
subtraction: the first row total is 120 (all the persons who own a CD
player, whether or not they own a car); the first column total is 90
(all the persons who own a car, whether they own a CD player or not).
It follows that the second row total is 180 (since both rows must total
300) and the second column total is 210 (likewise), so there are 120
persons who don't own a car AND don't own a CD player. (These are the
same 120 persons you could have discovered by subtracting the 180 who
owned one or the other or both from the 300.)
You might prefer to have percentages in the table, since these
are directly convertible into probabilities:
Own a Don't own
car a car TOTAL
Own a CD player: 10 30 40
Don't own a CD player: 20 40 60
TOTAL 30 70 100
> We will use the C to denote that the student owns a car and D to
> represent that the student owns a CD player.
Using this notation, you could have labelled the first column "C", the
second column "not-C", the first row "D", the second row "not-D".
> The events "owning a car" and "owning a CD player" are not mutually
> exclusive. A student may own both. Therefore
>
> P(C or D) = P(C) + P(D) - P(C and D)
>
> NOTE: This I part I understand.
>
> >From sample data we find
>
> P(C or D) = 180/300 = 0.60 P(C) = 0.30 P(D) = 0.40
>
> How on earth do they get P(C) = 0.30 P(D) = 0.40 ????
Well, you saw that P(C or D) = 0.60; you were told that
P(C but not D) = 20% = 0.20, and that P(D but not C) = 30% = 0.30;
and it followed that P(C and D) = 0.60 - 0.20 - 0.30 = 0.10.
This line of reasoning, of course, requires one to find
P(C and D) before knowing that P(C) = 0.30 and P(D) = 0.40).
Another line of reasoning observes that 300-180 = 120, which is
40% of 300, and this is P(neither C nor D). (This is the
lower-right-hand cell in the table.) Then P(not-C) = 0.30 + 0.40 = 0.70
and then P(C) = 1.00 - 0.70 = 0.30. Similarly,
P(not-D) = 0.20 + 0.40 = 0.60, and P(D) = 1.00 - 0.60 = 0.40.
Basically, however, what you need to do in cases like this is
to set up the 2x2 table, carefully figure out what the information
available implies about values in the table (either as counts or as
proportions [or percentages]), and calculate the remaining values.
There are only four independent values in the entire table.
> It continues ...
>
> Putting these in the formula above, we get
>
> 0.60 = 0.30 + 0.40 - P(C and D)
>
> So
>
> P(C and D) = 0.30 + 0.40 - 0.60 = 0.10
>
> The probability is 10% that a randomly selected student from the group
> owns both a car and a CD player.
>
> I have no idea where they get the numbers from.
------------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED]
MSC #29, Plymouth, NH 03264 603-535-2597
184 Nashua Road, Bedford, NH 03110 603-471-7128
