Homework # 3 - Math 2236
The problems on this homework must be done alone. The honor code is in effect.
You may submit the solutions twice, if you wish.
First Name: Last Name:
Questions 1 - 7 deal with the following set of logic gates.
In this image, a and b are the imputs and q is the output. Each is a boolean variable and can take the values 0/F or 1/T. You can see the line from a splits so that it goes into both the AND gate and the top NOT gate. And the line from B splits, too. The place where the line from A bends over the line from B means that there, the connections don't touch.
1. In a deck of 52 cards, how many 5 card hands are 2 pair (that means 2 cards of one value, two of another value, and then a fifth card that is different than either of the pairs)?
a. 123,552
b. 247,104
c. 267,696
d. 292,032
e. 658,944
2. What is the probability of being dealt a 5 card hand of exactly 2 pair from a well shuffled deck of 52 cards?
a. 2.75\%
b. 4.75\%
c. 6.25\%
d. 9.50\%
e. 11.25\%
3. Which of these is congruent to 5+6 (mod 4)?
a. 0
b. 1
c. 2
d. 3
4. Which of these is congruent to 5\(\times\)6 (mod 4)?
5. Consider the 3 hat problem, but change the number of people to 2, where each is given a hat of 3 colors. In how many color combinations can the hats be distributed?
a. 1
b. 3
c. 6
d. 9
e. 12
6. In the variation given in question 5 with 3 hat colors and a team of 2 players, how many differnt color combinations can each player see when they look at their partner (don't over-think this one)?
7. Can the variation given in question 5 with 3 hat colors and a team of 2 players be won 100\% of the time with the appropriate strategy?
a. yes
b. no
8. If we change it to 2 hat colors and 2 players, can this variation be won 100\% of the time with the appropriate strategy? (Okay, I'm going to tell you the answer to this question. It's yes. And you can figure out the strategy. It's much easier than the version we did in class. Remember, to make it work, the two players can never be right at the same time.)