Question

Assume that there are 6 different issues of Time, 7 different issues of Sports Illustrated, and 4 different issues of Popular Science, including the December 1st issue, on the rack. You choose 4 of them at random.

What is the probability that you choose the December 1st issue of Popular Science?

Answer 1

Total number of issues: 18
December 1rst Issue: Only One
Popular Science: 4

The chance of picking December 1rst issue is 1/5

Answer 2

that is wrong 1/5

Answer 3

We have a total amount of outcomes at:
\[
6+7+4=17
\]Thus, let us compute the chance of NOT picking the Popular Science magazine needed (since each event is independent of one another):
\[
P(E_1)P(E_2)P(E_3)=\frac{16}{17}\cdot\frac{15}{16}\cdot\frac{14}{15}\cdot\frac{13}{14}=\\
\frac{_{16}P_4}{_{17}P_4}=\frac{13}{17}
\]Therefore the probability of picking the December 1st edition is:
\[
1-P(E_n)=1-\frac{13}{17}=\frac{4}{17}
\]And we're done.

Answer 4

Using combinations, there are C(1,1)]*[C(16,3)] ways to choose the four magazines with the Dec 1 issue being selected [C(1,1)] which leaves C(16,3)] ways to choose the other three magazines.
The number of ways to choose 4 magazines from a group of 17 is [C(17,4)].

The probability of selecting 4 magazines with the Dec 1 issue as one of the three is the following:

C(1,1)]*[C(16,3)]/[C(17,4)] = 4/17.

Answer 5

ahh I misread.