Selected Answers to Problem Set 1

3. A general remark about genetics and probability: Indeed, because you inherit on the average 50% of your mother's (or father's) genes, you are 50% related to her/him. Your sibling will also inherit 50% of your mother's genes, meaning that on the average you share 25%. However, your sibling will also inherit 50% of your father's genes, meaning that you share another 25%, 50% in total. Answering questions (a) and (b) is done in this way; in both cases, you are equally related to each person in the separate conditions.

(c) Genetically, you are equally likely to be closely related to your aunt and to your brother's daughter. An easy way to look at this problem is to notice that you and your aunt share the same relationship as you and your brother's daughter. Genetically speaking, you share 50% of the genes of your mother/father, who in turn shares 50% of the genes with her/his sister. You also share 50% of the genes with your brother, who in turn shares 50% of his genes with his daughter. In both cases you are 25% similar to the persons in question.

(d) Genetically, you are more likely to be closely related to your uncle than to your first cousin's son. Similar to last problem, you share 25% of your genes with your uncle. As for your first cousin's son, your first cousin's father or mother is your uncle or aunt. Therefore, you share 25% with that specific aunt or uncle. That person's offspring will share 50% of that person, or 12.5%. That offspring's son, your first cousin's son, will share 50% once again, or 6.25%. Therefore, you share 6.25% of your genes with your first cousin's son which is much less than you share with your uncle.

4. The odds of NOT tossing aces (1's) once is 35/36. The odds of not tossing aces twice is (35/36)2. Three times is (35/36)3 and so on. We want n when (35/36)n > 0.5. 
n > ln(0.5) / ln(35/36)
  > (-0.693) / (-0.0282)
  > 24.6
n should therefore be 25.

If you got 18 or 19 as your answer, you went the wrong way. Instead of calculating the probability of NOT rolling aces, you calculated the probability OF rolling aces. In reality you found the average number of rolls of NOT rolliing aces. The answer is quite different from the 25.

5. Start with a simpler problem. With two people, the probability that they don't have the same code is 99.999%. Therefore the odds that they do is 1 - 99.999, or 0.001%. With three people, the probability that they don't are (9999/10000) * (9998/10000), or 99.970%. The probability that they do is 1 - 99.970, or 0.030%. With 100 people, the probability is 1 - (9999/10000) * (9998/10000) * ... * (9901/10000). The answer you get is 0.6086, which is the probability that two people do not have the same code. Therefore, the probability that they do have the same code is 39.14%. With 150 people, simply change the 100 in the Maple code to 150 and repeat. Maple returns 0.3253, and taking one minus this result yields 67.47%.

Note: Notice the small difference between this problem and problem 4. The dice problem utilized probability with replacement, the second probability without replacement. Be aware of how the problems differ and how to calculate the appropriate formulas in each.

7. Basically there are four groups in this class: male smokers, male non-smokers, female smokers and female non-smokers. We want to find the fraction of female smokers to total smokers, which is the sum of both male and female smokers. We can apply Bayes Theorem to this problem. 
P(f|s) = P(f)P(s|f) / P(s)
  P(s) = P(f)P(s|f) + P(m)P(s|m)

P(f|s) = P(f)P(s|f) / [P(f)P(s|f) + P(m)P(s|m)]
       = (1/3)(1/2) / [(1/3)(1/2) + (2/3)(1/3)]
       = (1/6) / [(1/6) + (2/9)]
       = (3/7)
Important! Remember to differentiate between P(f|s) and P(s|f)! The first is the probability that a person is a female, given that they are a smoker. The second is the probability that a person is a smoker, given that they are female. In math, the '|' restricts the set to a specific quality. In this case, P(f|s) restricts the set of people to just smokers. This symbol is used frequently in, you guessed it, set theory.
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