14.4: Chapter Exercises

Chapter Exercises

Exercises are included in most of the sections. Here we present some more difficult problems, extras for experts, although you now know enough to work at least some of the problems here. Answers to some of them are given below but think about the problems before looking (you will need a pretty good calculator to get the exact numbers; if you don’t have one, just work out the formulas).

1. The probability that you will get a car for graduation is 1/3, and the probability that you will get a new computer is 1/5, but you certainly won’t get both. What is the probability that you will get one or the other?
2. You have a 35% chance of getting an A in Critical Reasoning and a 40% chance of getting an A in Sociology. Does it matter whether the two outcomes are independent when you want to calculate the probability of at least one A? Does it matter whether the two outcomes are independent when you want to calculate the probability getting an A in both courses? Are they likely to be independent? Why?
3. Five of the 20 apples in the crate are rotten. If you pull out two at random, not replacing them as you pull them out, what’s the probability that both will be rotten?

The remaining problems are harder.

4. Now solve The Chevalier de Méré’s second problem. What is the probability of rolling at least one double six in twenty-four rolls of a pair of dice? Use the same strategy that was used above to solve his first problem.
5. Aces and Kings. Remove all the cards except the aces and kings from a deck. This leaves you with an eight-card deck: four aces and four kings. From this deck, deal two cards to a friend.
   1. If they look at their cards and tell you (truthfully) that their hand contains an ace, what is the probability that both of their cards are aces?
   2. If they instead tell you (truthfully) that one of their cards is the ace of spades, what is the probability that both of their cards are aces? The probabilities in the two cases are not the same.
6. It’s in the Bag. There are two opaque bags in front of you. One contains two twenty-dollar bills and the other contains one twenty-dollar bill and one five-dollar bill. You reach into one of the bags and pull out a twenty. What is the probability that the other bill in that bag is also a twenty?
7. The Monty Hall Problem. There are three doors in front of you. There is nothing worth having behind two of them, but there is a suitcase containing $50,000 behind the third. If you pick the right door, the money is yours. Pick a door: 1, 2, or 3. You choose door number 1. But before Monty Hall shows you what is behind that door, he opens one of the other two doors, picking one he knows has nothing behind it. Suppose he opens door number 2. This takes 2 out of the running, so the only question now is about door 1 and door 3. You may now reconsider your earlier choice: you can either stick with door 1 or switch to door 3.
   1. What is the probability that the money is behind door 1?
   2. What is the probability that the money is behind door 3?
   3. Do your chances of winning improve if you switch?
8. The Birthday Problem. How many people would need to be in a room for there to be a probability of 5 that two of them have a common birthday (born on the same day of the month, but not necessarily of the same year)? Assume that a person is just as likely to be born on any one day as another and ignore leap years.

Hint: Much as in the previous problem, it is easiest to use the rule for negations in answering this.

Answer

7. The Monty Hall Problem. We will work the answer out in a later chapter using the rules for calculating probabilities. For now, here are three hints (don’t look at the third until you have tried working the problem). First, you do improve your chances by switching to door 3. Second, think about what would happen if you repeated this process a hundred times. Third, draw a diagram representing all the things that could happen and not how often switching pays off compared to the total number of outcomes.
8. The Birthday Problem. The negation of the claim that at least two people in the room share a birthday is the claim that none of them share a birthday. If we can calculate the latter, we can subtract it from 1 to get the former.

Order the people by age. The youngest person was born on one of the 365 days of the year. Now go to the next person. They could have been born on any of the 365 days of the year, so the probability that their birthday differs from that of the first person is 364/365. Now move on to the next person. The probability that their birthday differs from those of the first and the second is 363/365. For the next person, the relevant probability is 362/365, and so on.

The birthdays are independent of one another, so the probability that the first four people have different birthdays is 365/365 x 364/365 x 363/365 x 362/365. There is a pattern here that we can generalize. The probability that the first N people have different birthdays is 365 x 364 x….x (365 - (N+1))/365N. And so, the probability that at least two out of N people have a common birthday is one minus all of this, i.e., (1 - (365 x 364 x….x (365- (N+1)))/365N. Now that we have this formula, we can see what values it gives for different numbers of people (and so for different values of N). When there are twenty-two people in the room N is 22, and the formula tells us that the probability that at least two of them have a common birthday is about .47. For twenty-three people, it is slightly more than a half (.507). For thirty-two people, the probability of a common birthday is over .75, and for fifty people it is .97. And with one hundred people there is only about one chance in three million that none share a common birthday.