# Probability Issue: Hints and Answers

Remember the Math Adventurer’s Rule: Figure it out for yourself! Whenever I give a problem in an Alexandria Jones story, I will try to post the answer soon afterward. But don’t peek! If I tell you the answer, you miss out on the fun of solving the puzzle. So if you haven’t worked these problems yet, go back to the original posts. If you’re stuck, read the hints. Then go back and try again. Figure them out for yourself — and then check the answers just to prove that you got them right.

This post offers hints and answers to puzzles from these blog posts:

## First, a Few Hints

Introduction to Probability: Monopoly

• To find the probability, make a fraction. The numerator is the number of possible successes: How many ways could you throw a total of 5 with two dice? The denominator is the total number of possible outcomes: How many ways might the dice land? You can read the probability fraction as a ratio or change it to a percent. For example, a probability of  $\frac{3}{4}$  could be read, “Three out of four,” or could be changed to 75%. Unfortunately, the answer to this problem isn’t anywhere close to  $\frac{3}{4}$ !

Alex’s Birthday Surprise

• Answer Dr. Jones’s question with the Problem Solving Tool of “Guess and check.” If you put Leon’s guess of n = 15 into the formula, you get a probability P = 0.98, which is 98%. That’s way too high. Check a few smaller numbers and narrow it down to the n that gives you about P = 0.5, or 50%. (It won’t be exactly 50%. In real life, probabilities rarely come out as simple fractions.)
• The second question is easy, with one minor trick: Count the dates carefully. Do not use the formula, because it only applies when you don’t care which birthdays match — but in this case, we are trying to match specific days. If you aren’t sure how to figure probability, see the hint for the Monopoly problem, above.
• The third question is much more difficult, but if you read A mathematician’s guide to birthdays, that will give you a start. In the same way that he calculates the probability of sharing a birthday out of the whole year (365 days), Dr. Jones’s formula calculates it for d days. First, find the chance of not sharing any birthdays, then subtract that from one.

Baby Blue Eyes

• Does the baby know what genes Alex has? Or perhaps a better way to say it: Does any given set of sperm and egg know how any other set of sperm and egg are made? To find the probability of two independent events both happening, multiply the probabilities of each individual event.

## Story Problem Challenge Hints

• Princess Kitten: The Dogs’ Adventure
Try the Problem Solving Tool of “Act the problem out, step by step.”
• One of the Three Musketeers: The Baker’s Predicament
Can Mrs. Sterns buy only part of a bag of chocolate chips?
• Horsey Girl: What Number Am I?
Try the Problem Solving Tool of “Make a systematic list,” and then cross out the numbers that don’t fit the clues.
• A Musketeer’s Sister: Ballerina Brainteaser
Try word algebra to help you think it through.
• Computergeek: A Ferengi at Heart
Watch out: Some of the numbers are given in cents, and some are dollars — don’t get confused! The $27 is your net profit, after paying for all the cans, even the ones that you didn’t sell. • The Engineer: An International Milk Mystery The three small bags are the same size, so the milk is split evenly. They don’t each hold a liter. • The Math Teacher: Homeschool Workshop Be careful: The problem doesn’t say there were 72 women! Try a bar diagram, like in Hobbit Math. ## The Answers Introduction to Probability: Monopoly • There are 6 possibilities for each die, which means 6 $\times$ 6 = 36 ways the dice may land. Only 4 of these possibilities let you move to Reading Railroad, so the probability is $\frac{4}{36} = \frac{1}{9}$ . One out of nine, or 11%. Alex’s Birthday Surprise • With a group of 7 people, all born in the same month, the odds are 50-50 that there will be at least one shared birthday. You may have noticed that our list of 7 mathematicians included a shared birthday. • There are 31 days in the month of December. Assuming that all days are equally likely (which is probably not an accurate assumption, but what else can we do?), the probability that a baby born in December would be born on one of the listed days is $\frac{6}{31}$ , or 19%. • Assuming that all the possible birthdays are equally likely, the probability of not sharing any birthdays is a fraction. The denominator is the all the different possible arrangements of birthdays. Since each of our n people could have ANY of the d days as a birthday, the possible arrangements are: $d \times d \times d \times ... = d^n$ The numerator of our probability fraction is a permutation: How many ways can we arrange n different dates, if we have d options to choose from? $_{d} P _{n} = \frac{d!}{\left ( d - n \right )!}$ So the chance of not sharing any birthdays is: $\frac {d!}{d^n \left( d-n \right)!}$ And finally, since there are only two options (either no shared birthdays or at least one shared birthday), we subtract that fraction from one to get Dr. Jones’s probability formula: $P = 1 - \frac {d!}{d^n \left( d-n \right)!}$ Baby Blue Eyes • Each baby is independent. No matter how many blue-eyed babies they’ve already had, the chance of the next baby having blue eyes is $\frac{1}{4}$, or 25%. Beware: You cannot multiply percents directly. You have to use fractions or decimals to multiply! If the chance of one blue-eyed child is 25%, then the probability of four children all having blue eyes is $\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{256}$ = 0.4%, or 1 out of 250. [Oops! See comments on the original post. I forgot to take into account that the family might have more than four children — but I’m not sure how to weight the calculation for various-sized families, so we’ll let this answer stand.] • But what about the parents? If we assume [by the Law of Large Numbers See comments below. I guess it’s time to reiterate my confession…] that the Punnett square represents the proportion of everybody’s genes in our population, then there are two Bb people out of every three people with brown eyes, which would mean that the probability of a brown-eyed person being able to have a blue-eyed child is $\frac{2}{3}$ , or about 67%. The chance of two brown-eyed parents both having Bb genes is $\frac{2}{3} \times \frac{2}{3}$, or about 44%. • Finally, the chance of these parents and children all coming together in one family would be $\frac{1}{256} \times \frac{4}{9}$ = about 0.2%, or 1 out of 500 — not 1 out of every 500 families in all, but 1 out of every 500 families with brown-eyed parents and four children. ## Story Problem Challenge Answers • Chickenfoot: How Much Money?$12.
• Princess Kitten: The Dogs’ Adventure
Each real dog has a matching stuffed animal. 10 dogs in all.
• One of the Three Musketeers: The Baker’s Predicament
63 + 52 = 115 ounces needed, which would be 5.75 bags. Mrs. Stern should buy 6 bags.
• Horsey Girl: What Number Am I?
9.
• A Musketeer’s Sister: Ballerina Brainteaser
8 counts = 1 pirouette + 2 fouettés.
1 pirouette = 4 counts.
1 fouetté = 2 counts.
1 pirouette + 32 fouettés = 4 + 32 $\times$ 2 = 68 counts.
• Computergeek: A Ferengi at Heart
You paid 75 $\times$ 25¢ = $18.75 for the cans of pineapple soda. That means your gross earnings were$18.75 + about $27 = about$45.75. You sold 46 cans, so lets round the gross earnings to $46. The cans of soda must have sold for$1 each.
• The Engineer: An International Milk Mystery
4 liters divided into 3 bags =  $\frac{4}{3}$  = $1 \frac{1}{3}$ liters per bag.
• The Math Teacher: Homeschool Workshop
Draw a bar divided into 7ths, and then shade 2 of the sections. That represents the men. Shade 2 more sections, which are the number of women that would equal the men — so the remaining sections are the “extra” women. These are 3 sections, representing the 72 more women than men, which means:
1 unit = 72 ÷ 3 = 24.
7 units = 7 $\times$ 24 = 168 people altogether.

## To Be Continued…

Read all the posts from the July/August 1999 issue of my Mathematical Adventures of Alexandria Jones newsletter.

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## 2 thoughts on “Probability Issue: Hints and Answers”

1. Nice set of problems, and a fun introduction to probability!

I have one quibble with your answers, though: “But what about the parents? If we assume (by the Law of Large Numbers) that the Punnett square represents the proportion of everybody’s genes in our population…” is not a good assumption (nor does it relate to the law of large numbers). The Punnett square describes the proportion of offspring in each category given the parents’ genetics. It doesn’t tell you anything about how common B and b genes are in the population! That’s determined by evolution. You’ll see quite a different ratio of BB to Bb depending on lots of factors: in Africa it’s probably close to 1 while in Sweden it’s probably close to 0, as one example. And there’s no reason to expect the worldwide average to be anywhere near 1/3.

2. I was thinking that the American melting pot would produce a wide mix of genes, but I was careless and didn’t think about the fact that there would (of course) be BB-BB and BB-Bb couples mixed in as well, which would obviously change the proportions. Oh, well — perhaps it’s a good lesson in not trusting that the answers given are actually correct?