Lectures 5 and 6: Introduction to Probability theory

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The birthday problem:

• Events are possible outcomes of some random processes
  • Examples of events :
    • you pass this class
    • 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) )

Recall that there are four possible phenotypes for the ABO blood group: Type A, Type AB, Type B, and Type O.

Suppose Prob(Type A) = 0.2, Prob(Type B) = 0.3, Prob(Type) AB= 0.1. What is Prob(Type O)?

Since Pr(A) + Pr(B) +Pr(AB) + Pr(O) = 1
Pr(O) = 1- Pr(A) - Pr(B) - Pr(AB) = 1 - 0.2 + 0.3 + 0.1 = 0.4

Likewise, what is the probabilty that you are NOT type O?

Pr(Not O) = 1- Pr(Type 0) = 1-0.4 = 0.6

• 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

  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

Suppose you have a forensic DNA maker that has five alleles with the following frequecies:

Freq(A₁) = 0.1, Freq(A₂) = 0.2, Freq(A₃) = 0.2, Freq(A₄) = 0.4, Freq(A₅) = 0.1,

What the expected frequency of an A₅A₅ homozygote?

To get an A₅A₅, that individual must get an A₅ from its father and get an A₅ from its mother.

If we have random mating, then

Pr(A₅ from father AND A₅ from mother) = Pr(A₅)*Pr( A₅ from mother) = 0.1² = 0.01

What the expected frequency of an A₁A₂ heterozygote?

There are two ways for this to happen: A₁ from father and A₂ from mother or A₂ from father and A₁ from mother.

Pr(A₁ from father and A₂ from mother or A₂ from father and A₁ from mother)

= Pr(A₁ from father and A₂ from mother) + Pr( A₂ from father and A₁ from mother).

= Pr(A₁ from father)*(Pr A₂ from mother) + Pr( A₂ from father)*Pr( A₁ from mother).

= 0.1*0.2 + 0.2*0.2 = 2*0.1*0.2 = 0.04

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.

• 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.

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,
  

• Recall (from your high school days ) one feature about taking the logarithm (or log) of a number,

  log(x^A) = A*log(x)

• Thus, we can solve for k by taking logs to give
  

• 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?