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

Reader's Digest - 0.6
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

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
a) .25 read only Reader's Digest
b) would be .15+.05+.1 = .3
etc.

To answer these questions, we can apply the formula P(A or B) = P(A) + P(B) - P(A and B).

a) The probability that a household chosen at random subscribes to only Reader's Digest can be calculated as follows:

P(Reader's Digest only) = P(Reader's Digest) - P(Reader's Digest and Chatelaine) - P(Reader's Digest and Maclean's) + P(Reader's Digest, Chatelaine, and Maclean's)

P(Reader's Digest only) = 0.6 - 0.2 - 0.2 + 0.05
P(Reader's Digest only) = 0.25

Therefore, the probability that a household subscribes to only Reader's Digest is 0.25.

b) The probability that a household chosen at random subscribes to Chatelaine, Maclean's, or both can be calculated as follows:

P(Chatelaine or Maclean's or both) = P(Chatelaine) + P(Maclean's) - P(Chatelaine and Maclean's)

P(Chatelaine or Maclean's or both) = 0.5 + 0.4 - 0.15
P(Chatelaine or Maclean's or both) = 0.75

Therefore, the probability that a household subscribes to Chatelaine, Maclean's, or both is 0.75.

c) The probability that a household chosen at random subscribes to one magazine only can be calculated by summing the probabilities of subscribing to each magazine individually and subtracting the probabilities of subscribing to more than one magazine:

P(One magazine only) = P(Reader's Digest only) + P(Chatelaine only) + P(Maclean's only)

P(One magazine only) = (0.6 - 0.2 - 0.2 + 0.05) + (0.5 - 0.2 - 0.15 + 0.05) + (0.4 - 0.2 - 0.15 + 0.05)
P(One magazine only) = 0.8

Therefore, the probability that a household subscribes to one magazine only is 0.8.

To answer these questions, you'll need to use the formulas you learned correctly. Let's go through each question one by one and determine which formula to apply.

a) What is the probability that a household chosen at random subscribes to only Reader's Digest?

To find the probability of a household subscribing to only Reader's Digest, you need to subtract the probabilities of households subscribing to both Reader's Digest and either of the other two magazines from the probability of subscribing to Reader's Digest alone.

So the formula to apply is:
P(only Reader's Digest) = P(Reader's Digest) - P(Reader's Digest and Chatelaine) - P(Reader's Digest and Maclean's)

= 0.6 - 0.2 - 0.2

b) What is the probability that a household chosen at random subscribes to Chatelaine, Maclean's, or both?

To find the probability of subscribing to Chatelaine, Maclean's, or both, you can simply add the probabilities of subscribing to each magazine and subtract the probability of subscribing to all three magazines (since we don't want to count it twice).

So the formula to apply is:
P(Chatelaine or Maclean's) = P(Chatelaine) + P(Maclean's) - P(Chatelaine and Maclean's)

= 0.5 + 0.4 - 0.15

c) What is the probability that a household chosen at random subscribes to one magazine only?

To find the probability of subscribing to one magazine only, you need to subtract the probabilities of subscribing to more than one magazine from the probability of subscribing to any magazine.

So the formula to apply is:
P(one magazine only) = P(Reader's Digest) - P(Reader's Digest and Chatelaine) - P(Reader's Digest and Maclean's)
+ P(Chatelaine) - P(Reader's Digest and Chatelaine) - P(Chatelaine and Maclean's)
+ P(Maclean's) - P(Reader's Digest and Maclean's) - P(Chatelaine and Maclean's)

= 0.6 - 0.2 - 0.2 + 0.5 - 0.2 - 0.15 + 0.4 - 0.2 - 0.15

By applying the correct formulas to each question, you can calculate the desired probabilities.