# math - classifying events

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A market survey has estimate that the probability of a household's subscribing to the following magazines is:

Chatelaine - 0.5
Maclean's - 0.4
Reader's Digest and Chatelaine - 0.2
Reader's Digest and Maclean's - 0.2
Chatelaine and Maclean's - 0.15
All three - 0.05

a) What is the probability that a household chosen at random subscribes to only Reader's Digest?
b) What is the probability that a household chosen at random subscribes to Chatelaine, Maclean's, or both?
c) What is the probability that a household chosen at random subscribes to one magazine only?

Today, I learnt 3 new formulas..

P(A and B) = P(A) * P(B)
P(A or B) = P(A) + P(B)
P(A or B) = P(A) + P(B) - P(A and B)

and I'm not sure which one to apply to which question. =S

• math - classifying events -

make sure you understand that difference between the last two, since the left side is the same

P(A or B) = P(A) + P(B) - P(A and B) is the more general formula and used when there is an overlap in the events of A and B
If A and B are independent of each other, the last term drops off, because it has a value of zero.

There is an extension of this ...
P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

so P(A or B or C) = .6 + .5 + .4 - .2 - .2 - .15 + .05
= 1 , showing that everybody must have one or more of these magazines

I would do this question with Venn diagrams.
Draw 3 overlapping circles , R, M, and C for the magazines
fill in the overlap of all 3 with .05
now look at the intersection of R and M, which is .2
but we have already counted part of that with the .05, so put 15 in the remaining part.
Look at the overlap of R and C, which is .2, but .05 has already been counted, so put .15 in the open part.
Look at the overlap of M and C, which is .15, again .05 is already counted so put .1 in what is left.

Now look at the whole of circle R, which is .6 in total.
But we already have counted .15+.05+.15 or .35 leaving .25 for the empty part of R.
In the same way the empty part of M is .1 and that of C is .2

You can now just read off any answer you want, so for
b) would be .15+.05+.1 = .3
etc.

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