Solution: This question is almost exactly like the question on Quiz 6, and of course so is the solution.
One difference is that we want the probability of at least one pair, rather than the probability of no pairs (the other is that there are 7 cards, rather than 5). It is easiest to do this by computing the probability of no pairs, and then subtracting that result from one.
Notice that ``no pairs'' is the same as saying ``all the cards have different numbers''. So, we need to count how many ways we can pick 7 cards from a deck, all with different numbers. It is easiest if we view our hand as being ordered- that is, we distinguish the order in which we pick the cards.
First we pick the first card. There are 52 choices for this card.
For the second card, we have only 48 choices, since we cannot pick any cards which are the same number as the first. For the third card, we have 44 choices: we can't use the 4 cards that are the same number as the first card, nor the same number as the second.
Continuing in this way, we see that there are ways to pick 7 cards with no pairs, paying attention to which is first. (If we wanted to ignore the order, we would divide by ).
Since there are
different ways to pick 7 cards, the probability of no pairs is
Solution: You might think this is true since the probability of rolling one die and getting a 6 is , and . But probabilities only add like that for mutually exclusive events.
In this case, we can calculate the probability several ways. The
addition rule says that it is
Solution: For all the questions below, we will assume we can distinguish the dice, either because one is red and one is white, or we just roll one die and then the other. This means that there are always 36 different possible rolls (since a 3 on the first and a 6 on the second is different from a 6 on the first and a 3 on the second).
Solution: There is only one way to get a sum of two, by rolling two ones. Thus, the probability is .
Solution: Now we have to think a little more. We can get a sum of 7 in six different ways: , , , , , and . Thus, the probability is , that is .
Solution: To have a sum of 10 or more, we have six ways: , , , , , and . So the probability is , or .
Solution: If you know one is a 5, then the other one must be a 5 or a 6. The probability of rolling one die and getting a 5 or a 6 is , or .
You could use conditional probability to do this, if you really
Solution: The graph looks pretty much like the sidewalk itself:
Solution: No. To have an Eulerian circuit, all vertices must be even. But the graph in question has two odd vertices.
Solution: Yes. Since there are two odd vertices, any Eulerian path must start at one of them and end at the other. One such path is shown here.
|Kansas City||Omaha||St. Louis||Wichita|
Solution: Using the nearest neighbor algorithm starting from Witchita, we take the shortest edge, which is the one to Kansas City at 280 miles. From Kansas City, it is slightly shorter to Omaha (220 mi). From Omaha, we must go to St. Louis (310 mi) and then return to Witchita (500 mi). This is a total of 1310 miles.
Wichita, Kansas City, Omaha, St. Louis , Wichita. Distance: 1310 miles
Solution: For the greedy algorithm, rather than choosing the shortest path as we go, we choose the overall shortest paths, only ruling out those that will mess us up. So, the shortest leg of the journey is KC-Omaha (220 mi) followed by KC-St. Louis (225 mi). The next shortest is Witchita-KC, but we can't use that one and still use the other two (it makes a 3-way fork). So, instead we use the Witchita-Omaha segment (300 mi). Using this rules out using the Omaha-St. Louis road, so we have to complete the journey by taking the long road from St. Louis to Witchita (500 mi). This gives an overall journey of 1245 miles, slightly shorter than the nearest neighbor algorithm does.
Wichita, Omaha, Kansas City, St. Louis, Wichita. Distance: 1245 miles
Solution: The only way to find the shortest possible path for the travelling salesman problem is to check ALL possible round-trip routes. Fortunately, there is only one other one (since reversing the direction doesn't change the overall distance). This is the one that avoids the Witchita-St. Louis road, and has a total distance of 1115 miles.
Wichita, Omaha, Kansas City, St. Louis, Wichita. Distance: 1115 miles
Solution: If H moves to the center as indicated, V has no choice but to move on the end. H can then win by moving as shown, blocking any move of V's.
Note that if H plays the other possible opening move, V can respond with a winning move by placing his tile adjacent to H's.
Solution: V has two possible different opening moves, so we have to show H's response to each. In fact it doesn't matter what H does, but we show a possible response to each of V's moves. In both cases, V has no choice where to play following, and H always wins.
Solution: Since the payoff matrix is already reduced (Sorry, I goofed), there is no clear ``best stategy''. Probably the best bet here would be to use mixed strategy, where each candidate campains on domestic issues half the time and foreign policy the other half.
However, no one realized this was possible, so I pretty much accepted any reasonable choice with a decent explanation.
Solution: This depends on your answer to the previous part. Of course, your answer must be something between -2 and +4. If the mixed strategy outline above is followed, the democrat should expect to gain 2 points.
|Qatar H||Qatar L|
|Kuwait H||(9, 9)||(5, 17)|
|Kuwait L||(17, 5)||(6, 6)|
Assuming both Kuwait and Qatar act in order to maximize their own profit, and without communicating with each other, what action will each country take, and what is their anticipated revenue? Justify your answer.
Solution: Our first step is to reduce the payoff matrix.
Note that Kuwait always prefers low prices (L), since if Qatar charges high prices, they expect to make 17 million (instead of 9 million if Kuwait charges high), and if Qatar charges low prices, Kuwait makes 6 million instead of 5 million. So we can cross out the Kuwait H row, leaving us with
|Qatar H||Qatar L|
|Kuwait L||(17, 5)||(6, 6)|
Now Qatar also will charge low prices, since they will make 5 million if they charge high, but 6 million if they charge low. This reduces the payoff matrix to the single entry
|Kuwait L||(6, 6)|
Thus, both countries will charge low prices, and each expects to make 6 million.