Hello everyone.
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 ) :
We will use the C to denote that the student owns a car and D to
represent that the student owns a CD player.
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 ????
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.
If you could explain the solution, I would be extremely greatfull!!
Andrew
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Thanks
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