Lecture 14: Introduction to Probability theory

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• Events are possible outcomes of some random processes
  • Examples of events :
    • you pass 320
    • The genotype of a random individual is Bb
    • the weight of a random individual is less than 150 pounds

• We can define the probability of a particular event, say A, as the fraction of outcomes in which event A occurs.

• Denote Probability of A by Pr(A), or Prob(A)

• For example, when flipping a coin once, the possible outcome is heads or tails.
  • Pr(Head) = 0.75 means that chance is 75% that the coin will be a head and hence

  • Pr(Tail) = 1 - Pr(Head) = 0.25.

• Probabilities are between zero (never occur) and one (always occur)
  • Pr(A) lies between zero and one for all A.

• Probabilities sum to one
  • The sum of probabilities of all mutually exclusive events is one.
    • For example, if there are n possible outcomes, Pr(1) + Pr(2) + .. + Pr(n) = 1

    • Hence, Pr(1) = 1 - ( Pr(2) + .. + Pr(n) )

• AND rule: If A and B are independent events (knowledge of one event tells us nothing about the other event), then the probability that BOTH A and B occur is
  • Pr(A and B) = Pr(A) Pr(B)

  • Hence generally AND = multiply probabilities

• OR rule: If A and B are exclusive events (nonoverlapping), then the probability that EITHER A or B occurs is
  • Pr(A or B) = Pr(A) + Pr(B)

  • Hence generally OR = add probabilities
  

  Example 1

  Suppose we are rolling a fair dice and flipping a fair coin

• What is the probability of rolling an even number on the dice?
  • A single roll of a fair dice has possible outcomes 1, 2, 3, 4, 5, 6 each with the same probability, 1/6. Rolling an even number means rolling 2 OR 4 OR 6. These three events (2, 4, 6) are nonoverlapping, and hence exclusive, so we can use the OR = add rule, giving
    • Pr(Roll even) = Pr(2) + Pr(4) + Pr(6) = 3/6 = 1/2

• What is the probability of rolling a 5 and then getting a head in the coin flip?
  • The dice roll and coin flip are independent events as the outcome of one does not influence the outcome of the other. Hence,
    • Pr( Head AND roll 5) = Pr(Head) * Pr(5) = 1/2*1/6 = 1/12

How do we compute joint probabilities when A and B are NOT independent (i.e., knowing that A has occurred provides information on whether or not B has occurred).

• The joint probability of A and B, Pr(A,B) , is the product of the probability of B, Pr(B), with the Probability of A given B, Pr(A | B).
  • Pr(A,B) = Pr(A | B) Pr(B)

• Pr (A | B) is called the conditional probability of A given B
  • Pr (A | B)= Pr(A,B) / Pr(B)

• A and B are said to be independent if Pr(A | B) = Pr(A), so that knowing event B occurred gives us no information about event A.

• An important use for conditional probabilities is to compute the probability of some complex event by conditioning on other events. For example, suppose that event A occurs under one of three other (mutually exclusive) events, say B, C, and D. Then
  • Pr(A) = Pr(A|B)*Pr(B) + Pr(A|C)*Pr(C) + Pr(A|D)*Pr(D)

  For example, suppose there are three genotypes with different disease risks, where event A is having the disease, and B, C, and D are three different genotypes. Pr(A|D) is the risk of the disease for genotype D, and so forth. The overall risk of the disease is just the weighted risk over all genotypes.

• Suppose we cross two Aa parents, where AA and Aa offspring are yellow, while aa offspring are green. Here,
  • Pr(AA) = Pr(aa) = 1/4
  • Pr(Aa) = 1/2
  • Thus, Pr(Yellow) = Pr(AA) + Pr(Aa) = 3/4.
  What is the probability that a yellow offspring has genotype Aa?
  

  Using our formula,

  • Pr(genotype = Aa | offspring = Yellow) = Pr(genotype =Aa, offspring = Yellow) / Pr(offspring = Yellow) = (1/2)/(3/4) = 2/3.

  Likewise

  • Pr(genotype = AA | offspring = Yellow) = (1/4)/(3/4) = 1/3.

What is the risk that you will have a disease given your sib (brother/sister) does?

This is quantified by the disease relative risk, RR, where

• RR = Prob(sib 1 affected | sib 2 is) / Prob(random individual affected)
• Thus, RR is the increase in your risk over that for a random individual.
• Note that RR = 1 if Prob(sib 1 affected | sib 2 is) = Prob(random individual affected), i.e. you have no increased risk given a relative has the disease.

Hence the disease relative risk is the increase in the conditional probability for a sib (or other relative) vs. a random individual.

As an example, consider diabetes. The probability that a random individual (from the US population) has type 1 diabetes is 0.4 percent. This is also referred to as the population prevalence, K. However, the frequency of diabetes in families with an affected sib is 6 percent. The resulting relative risk that an individual has diabetes, given that its sib does, is 6/0.4 = 15.

What is the probability that a pair of sibs both have diabetes?

• Pr(Both sibs affected) = Pr(2nd affected | 1st is) Pr(1st affected) = 0.06 * 0.004 = 0.00024

• Note that Pr(2nd affected | 1st is) = RR*K, as RR = Pr(2nd affected | 1st is) / K. Hence Pr(Both sibs affected) = (RR*K)*K = (K²)* RR

• Hence, the population frequency of families with both sibs affected is 15 times more common than expected by chance (i.e., if the disease is independent of family membership, which is K²).

Consider the following data for individuals with rheumatoid arthritis (from Del Junco et al, 1984)

• Prob(2nd sib affected | 1st sib affected) = 21 / 496 = 0.042
• Prob(random affected) = 12 / 673 = 0.018
• Relative Risk, RR = 0.042 / 0018 = 2.374

Prob(win jackpot) = (6/40)*(5/39)*(4/38)*(3/37)*(2/36)*(1/35) = 1/ 5,245,786

Suppose you buy 100 different lotto tickets for each drawing. How many such drawings do you have to play to have a 50 percent chance of winning (at least) one jackpot?

• Since you pick 100 out of 5,245,786 possible numbers, the probability of winning on any given drawing is
  • 100 / 5,245,786 = 0.000019

• Likewise, the probability of losing is
  • 1- Pr( winning ) = 1 - 0.000019 = 0.999981.

• The probability of losing k drawings in a row is
  

• We are interested in the number of drawings k such that this probability is 0.5 or less,
  

• Solving for k by taking logs gives
  

• Lotto drawings are twice a week, so there are 52*2 = 104 drawings per year, taking you 36,360/104 = 363 years to have this many plays. If you win on the 36,360-th try, you will have spent almost $3.64 million to win (most likely) 1.5 million. See why the state of Arizona likes Lotto?