Math 3118 Exam 1 Solutions

(15 points) 1. Two fair six sided dice are rolled 7 times. Find the probability that a sum of 6 is shown exactly two times.

Solution: A sum of 6 occurs with prob p=5/36, and we have n=7 independent trials, and need k=2 successes. So the answer is

(7 choose 2)*(5/36)^2*(31/36)^5= about 19.2%.

(20 points) 2. Suppose that I offer you the following bet. Should you take it? We each put down one dollar. You draw 4 cards without replacement from a 52 card deck. If you draw exactly two red and exactly two black cards, you win.

Solution: Let's find the probability that you win. There are (52 choose 4) ways to draw the 4 cards. There are (26 choose 2)*(26 choose 2) ways for you to draw 2 black and 2 red cards. So you win (26 choose 2)*(26 choose 2)/(52 choose 4)= about 39% of the time, which is less than half of the time. Thus it is best that you don't bet with me.

(25 points) 3. Suppose that a fair coin is flipped 4 times. Let E be the event that the first flip is a head, and let F be the event that at least two tails appear. Find P(E), P(F), and P(E\cap F), and determine if E and F are independent events. Explain your answer.

Solution: To find P(E): There are 2*2*2*2=16 possible coin flips. 8 of them begin with H, so P(E)=8/16=1/2.

To find P(F): Of these 16 coin flips, (4 choose 2)=6 have exactly 2 T's, (4 choose 3)=4 have exactly 3 T's, and (3 choose 4)=1 have exactly 4 T's. So P(F)=11/16.

To find P(E \intersect F): These are flips which start with an H, and have at least 2 T's. So either two fo the last 3 flips are T ((3 choose 2)=3 choices), or all of the last 3 flips are T (1 choice). Thus P(E \intersect F)=4/16.

To see if E and F are independent, check if P(E \intersect F)=P(E)*P(F). Does 4/16=1/2*11/16? No, so they are dependent events.

(20 points) 4. In a certain math class 65% of the students pass, and 35% of the students fail. 70% of those that pass always attend the class, while 80% of those who fail do skip class sometimes. Find the probability that a student passes if she always attends the class.

Solution: P(PASS and ATTEND)=0.65*0.70=0.455=45.5%.

P(FAIL and ATTEND)=0.35*0.20=0.07=7%.

So P(PASS|ATTEND)=45.5/(45.5+7)=about 86.7%

(20 points) 5. Find two non-disjoint events E and F which are not equal such that P(E|F)=P(F|E). Explain your answer in complete sentences.

Solution: Since P(E|F)=P(E\intersect F)/P(F), and

P(F|E)=P(E\intersect F)/P(E), all we need to do is find events E and F which are not equal, do intersect, and have P(E)=P(F).

We could flip a fair coin twice and let E= first flip is H={HH,HT}, F= second flip is T={TT,HT}. Then P(E)=P(F)=1/2, E\intersect F={HT} is non-empty, and E does not equal F.