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Probability - Biometry - Lecture Notes | ANSC 5403, Study notes of Biometrics

Texas Tech University (TTU)Biometrics

Material Type: Notes; Class: Biometry; Subject: ANIMAL SCIENCE; University: Texas Tech University; Term: Unknown 2008;

Typology: Study notes

Pre 2010

Uploaded on 03/19/2009

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AnSc 5403
Biometry
Lecture Notes 5
I. Probability
A. Probability theory is fundamental to inferential statistics
1. Definition (from Remington and Schork, 1970) – If a procedure can result in n
equally likely outcomes, nA of which have the attribute A, the attribute A has the
probability nA/n
a. In equation form – P(A) = nA/n
2. Properties of probability
a. P(A) ≥ 0 for any attribute A. If nA = 0, P(A) = 0.
b. P(A) ≤ 1 for any attribute A. If all n outcomes are nA, P(A) = 1.
c. If B denotes the absence of the attribute A, then P(A) + P(B) = 1.
(a) Coin toss – Heads vs. Tails
3. Example 1 – Draw one card from a deck (52 cards with four suits and 13
denominations). What is the probability of drawing a card from any one suit?
a. P(S) = 13/52 = ¼ = 0.25
4. Example 2 – Two pennies are tossed together. What is the probability of getting two
heads?
a. What are the possible outcomes?
(a) HH, HT, TH, TT
(b) P(HH) = ¼ = 0.25
5. The probabilities computed in Examples 1 and 2 are referred to as – a priori
probabilities
a. Such probabilities can be computed without actually drawing the card or tossing
the coins
B. Determining n – calculating the possible combinations
1. Ordered sequences
a. If a k objects are to be arranged in order, how many possible arrangements are
there?
(a) There are k choices to fill the first position, k – 1 choices to fill the second
position once the first is filled, and so on
(i) The number of combinations = k! = k x (k - 1) x (k - 2)…(3) x (2) x (1)
1. This arrangement of objects in a particular order is referred to as a –
permutation of the objects
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AnSc 5403 Biometry Lecture Notes 5

I. Probability A. Probability theory is fundamental to inferential statistics

  1. Definition (from Remington and Schork, 1970) – If a procedure can result in n equally likely outcomes, n (^) A of which have the attribute A, the attribute A has the probability nA/n a. In equation form – P(A) = nA/n
  2. Properties of probability a. P(A) ≥ 0 for any attribute A. If nA = 0, P(A) = 0. b. P(A) ≤ 1 for any attribute A. If all n outcomes are nA, P(A) = 1. c. If B denotes the absence of the attribute A, then P(A) + P(B) = 1. (a) Coin toss – Heads vs. Tails
  3. Example 1 – Draw one card from a deck (52 cards with four suits and 13 denominations). What is the probability of drawing a card from any one suit? a. P(S) = 13/52 = ¼ = 0.
  4. Example 2 – Two pennies are tossed together. What is the probability of getting two heads? a. What are the possible outcomes? (a) HH, HT, TH, TT (b) P(HH) = ¼ = 0.
  5. The probabilities computed in Examples 1 and 2 are referred to as – a priori probabilities a. Such probabilities can be computed without actually drawing the card or tossing the coins B. Determining n – calculating the possible combinations
  6. Ordered sequences a. If a k objects are to be arranged in order, how many possible arrangements are there? (a) There are k choices to fill the first position, k – 1 choices to fill the second position once the first is filled, and so on (i) The number of combinations = k! = k x (k - 1) x (k - 2)…(3) x (2) x (1)
  7. This arrangement of objects in a particular order is referred to as a – permutation of the objects
  1. Example – Four top students in the college all have the same GPA – a committee is chosen to rank the students after an oral interview. How many possible combinations of ranked orders are there among the four students? a. 4 x 3 x 2 x 1 = 24 possible combinations b. Alternatively, if we have k objects, but only wish to order r of them in all possible ways, how do we calculate the number of combinations? (a) (^) k k_Permutations_r = k!/(k-r)! (b) Example – Suppose we have five top students in the college, all with the same GPA, but the committee decides to randomly drop two of them from consideration for the top student award (i) 5!/(5 – 3)! = 5 x 4 x 3 x 2 x 1/(2 x 1) = 60 combinations
  2. Combinations – sets of objects without regard to order a. If we want to determine the number of ways in which r objects might be selected from a larger group of k objects, regardless of the order, how many ways are there? (a) We have five objects, designated A, B, C, D, and E (i) If order is ignored, how many possible combinations of two objects are there?
  3. Answer = AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
  4. Formula = k_Combinations_r = k!/[(k-r)!r!]
  5. Distinguishable permutations of objects not all different a. Consider the word BABA – one cannot distinguish the B’s and A’s (a) Distinguishable permutations of BABA = BABA, BAAB, BBAA, ABAB, ABBA, and AABB (b) For BABA, the total number of objects (k) = 4 (i) There are 2 (k 1 ) B’s and 2 (k 2 ) A’s (c) Formula – Number of distinguishable permutations of objects not all different = k!/k 1 !k 2 !...k (^) t !, where t = the kt equals the last possible kind of object C. Probability of composite events
  6. Mutually independent (mutually exclusive) events a. Suppose a process can result in n equally likely events, nA of which have attribute A, and nB of which have attribute B b. As before, P(A) = n (^) A/n and P(B) = nB /n c. Suppose the process is a coin toss. What is the probability of getting two heads if two tosses of the coin are performed?

(i) The usual form of describing these probabilities is to use the notation P (A|B) – “the probability of A given B.” Note that B cannot equal 0. (b) P(4-yr-old|cow) = 10/50 = 0. (c) Notice that this differs from the P(cow|4-yr-old) = 10/30 = 1/3 = 0. (d) General formula – P(A|B) = P(A and B)/P(B) E. Bayes’s Rule

  1. Example (from Remington and Schork, 1970) a. The prevalence of a particular disease in women is 1% = P(D) (a) The probability of the absence of the disease P(Da) = 1 – P(D) b. A new test has been found to give a positive result in 95% of women who have the disease (a) Thus, P(S|D) = 0. c. The test gives a false positive result (positive when given to women who do not have the disease) in 3% of women (a) Thus, P(S|Da) = 0. d. Suppose we want to know the probability that a woman has the disease given that the test result is positive – P(D|S) e. Bayes’s rule – P(D|S) = P(D) x P(S|D)/P(D) x P(S|D) +P(Da) x P(S|Da) (a) P(D|S) = (0.01 x 0.95)/[(0.01 x 0.95) + (0.99 x 0.03)] = 0. (b) Thus, we would expect 24% of women who test positive for the disease to actually have the disease