(1) Suppose that A and B are subsets of a universal set U, and that n( U ) = 50, n(A) = 10, n(A ∪ B) = 20, and n(A ∩ B) = 3. Supply the following information.

(a) n( A' ) = 40

(b) n( B ) = 13

(c) n( A' ∪ B ) = 43

(d) n( A' ∩ B ) = 10

  _________________________________________ U _____________
  |  ______________________                               |
  |  |                    |                          30   |
  |  |            ________|______________________         |
  |  |      7     |       |                     |         |
  |  |            |    3  |                     |         |
  |  |            |       |                     |         |
  |  |___  A _____|_______|        10           |         |
  |               |                             |         |
  |               |                             |         |
  |               |                             |         |
  |               |_____________ B _____________|         |
  |                                                       |
  |_______________________________________________________|

(2)

(a) In how many ways can you choose a committee of 4 people from a group of 10 people?

C_{10, 4} = 10! / (4!(10-4)!) = 10! / (4! 6!) = 210

(b) From a group of 10 people, you select a president, a vice-president, a secretary and a treasurer. In how many ways is this possible?

P_{10, 4} = 10! / (10-4)! = 10! / 6! = 5040.

(3) Suppose A and B are events, and P(A) = 1/2, P(B) = 1/3, and P(A ∪ B) = 2/3.
Are A and B independent, dependent, or is there not enough information given to say for sure? Explain.

We would like to see if P(A ∩ B) = P(A)P(B), but we don't have a value for P(A ∩ B).

However, this value can be obtained by the addition formula:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
2/3 = 1/2 + 1/3 - P(A ∩ B)
P(A ∩ B) = 1/2 + 1/3 - 2/3 = 1/6

Thus P(A ∩ B) = 1/6,
while P(A)P(B) = (1/2)(1/3) = 1/6.

Conclusion: P(A ∩ B) = P(A)P(B), so A and B are independent.

(4) One card is drawn off a 52-card deck. What is the probability that it is...

(a) a heart or a King? 16 cards are a heart or a King, so the probability is 16/52 = 4/13 ~ 30.769%
Alternative: let A = "Heart, " B = "King," then use formula to compute P(A ∪ B).

(b) a heart and a King? Only one card meets this description, so probability is 1/52 ~1.92%
Alternative: use formula for P(A ∪ B).

(c) neither a heart nor a King? There are 36 cards that are neither hearts nor Kings, so the probability is 36/52 = 9/13 ~ 69.23%
Alternative: combine formula for complement with your answer from part (a).

(d) a heart, given that it's also a King? If it's a King, then the chance of its being a heart is just 1/4 = 25%
Alternative: use formula for conditional probability.

(a) the first 2 tosses are heads?
To begin, all parts of this problem will involve the same sample space, so let's consider that sample space.
A typical simple event in the sample space will look like HHTHTT, i.e. just a sequecne of six slots filled with H's and T's.
By the multiplication principle n(S) = 2⁶ = 64.

Now let E = "first two tosses are heads."
Then simple events in E look like HH_ _ _ _, where the last 4 slots can be filled in with either H's or T's.
Thus by the multiplication principle, n(E) = 2⁴ = 16.

Now the answer to part a is n(E)/n(S) = 16/64 = 1/4 = 25%

(b) exactly 2 of the 6 tosses are heads?
Let F be the event "exactly 2 tosses are heads."
To make a simple event in F, you first would select 2 of six slots for H, and then fill in the remaining four slots with T.
For example, first choose the two slots _ _ H _H _ , then fill in TTHTHT.

Thus the number of simple events in F equals the number of ways to choose 2 slots from 6, namely
n(F) = C_{6, 2} = 15.

Conclusion: P(F) = n(F)/n(S) = 15/64 ~ 23.43%

(c) less than 2 of the 6 tosses are heads?
This means no heads or just 1 head. Call this event E.
Then E = {TTTTTT, HTTTTT, THTTTT, TTHTTT, TTTHTT, TTTTHT, TTTTTH}
and P(E) = n(E)/n(S) = 7/64 ~ 10.93%

(d) the first 2 tosses are heads or the last toss is a head?
Let A be the event "first 2 tosses are heads."
Let B be the event "last toss is a head."
We seek P(A or B) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B) =
n(A)/n(S) + n(B)/n(S) - n(A ∩ B)/n(S) =
16/64 + 32/64 - 8/64 = 40/64 = 5/8 = 62.5%

(6) At a certain college, 40% of the students are male, and 60% are female. Also, 20% of the males are smokers, and 10% of the females are smokers.

                      
                 s   P(ms) = (.4)(.2) = .08
                /
             .2/
              /
             /
            m
           / \
       .4 /   \
         /   .8\
        /       \
       /         n   P(mn) = (.4)(.8) = .32
       \         s   P(fs) = (.6)(.1) = .06
        \       /
         \   .1/
       .6 \   /
           \ /
            f
             \
              \
            .9 \
                \
                 n  P(fn) = (.6)(.9) = .54

(a) A student is chosen at random. What is the probability that the student is a male nonsmoker?

The sample space is S = {ms, mn, fs, fn}.
(where m=male, f=female, s=smoker, n=nonsomker)

By the probability tree above, we see that P(mn) = .32 = 32%.

(b) A student is chosen at random. If the student is a smoker, what is the probability that the student is female?

Let F = {fs, fn} = "student is female."
Let M = {ms, mn} = "student is male."
Let S = {ms, fs} = "student is smoker."

We are looking for P(F|S). According to Baye's Formula,
P(F|S) = P(F∩ S) / ( P(F∩ S) + P(M ∩ S))
= P(fs) / ( P(fs) + P(ms) ) = .06 / (0.6 + .08) = .06/.14 = 3/7 ~ 42.857%