# Definition and Calculus of Probability

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3 In experiments with multivariate outcome variable, knowledge of the value of one variable may help predict another. For now, the word prediction will mean update the probabilities of events regarding the other variable. The updated probabilities are called conditional probabilities. For example, 1. Knowing a man s height helps update the probability that he weighs over 170lb. 2. Knowing which assembly line a product came from, helps update the probability that it has a particular defect. 3. Knowing a person s education level helps update the probability of that person being in a certain income category.

4 Given partial information about the outcome of the experiment, results in a reduction of either the sample space or the number of eligible units. For example: Example If the outcome of rolling a die is known to be even, what is the probability it is a 2? If the selected card from a deck is known to be a figure card, what is the probability it is a king? Given event A = {household subscribes to paper 1}, what is the probability of B = {household subscribes to paper 2}? P(A B ) P(A B) A B U P(A B) U U

5 Definition The conditional probability of the event A given the information that event B has occurred is denoted by P(A B) and equals P(A B) = P(A B), provided P(B) > 0 P(B) Note that P(B B) = 1, which highlights the fact that when we are given the information that B occurred, B becomes the new sample space. Proposition The set function P( B) satisfies the three axioms of probability.

6 Example In the card game of bridge, the 52 cards are dealt out equally to 4 players called East, West, North and South. Given that North and South have a total of 8 spades among them, what is the probability that East has 3 of the remaining 5 spades? Use the reduced sample space defined by the given information. Solution: In the reduced sample space experiment there are 26 cards containing 5 spades and will be divided randomly between East and West. Thus, the probability that East ends up with 3 of the remaining 5 spaces is ( 5 21 ) 3)( 10 ( 26 ) 13

7 Do each of the following examples in two ways: a) by considering the reduced sample space, and b) by applying the formula for the conditional probability. Example 1. Roll a die twice. Given that the first roll results in 3 find the probability the sum is Roll a die twice. Given that one of the roll results in 3 find the probability the sum is An urn contains r red and b blue balls. n balls (n r + b) are selected at random and without replacement. Given that k of the n balls are blue, what is the probability that the first ball chosen is blue?

8 The Multiplication Rule The definition of P(A B) yields the formula: P(A B) = P(A B)P(B) or P(A B) = P(B A)P(A) The rule extends to more than two events. For example, P(A B C) = P(A)P(B A)P(C A B)

9 The Multiplication Rule: EXAMPLES 1. 40% of bean seeds come from supplier A and 60% come from supplier B. Seeds from supplier A have 50% germination rate while those from supplier B have a 75% rate. What is the probability that a randomly selected seed came from supplier A and will germinate? ANSWER: P(A G) = P(G A)P(A) = = Three players are dealt a card in succession. What is the probability that the 1st gets an ace, the 2nd gets a king, and the 3rd gets a queen? ANSWER: P(A B C) = P(A)P(B A)P(C A B) = =

10 Example An urn contains 8 red, each having weight r, and 4 white balls, each having weight w. Two are selected without replacement. What is the probability both are red if at each draw: 1. each ball is equally likely, and 2. the probability of each ball equals its weight divided by the sum of weights of all balls currently in the urn.

11 Example (The Mens Hats Problem) In the Men s Hats Problem, find the probability that exactly k of the N men get their own hat. Solution: Let F i = {man i gets his own hat}, E = {men 1,..., k get their own hats} = F 1 F k, and G = {men k , N do not get their own hats}. Then, P(E G) = P(E)P(G E) and P(E) = P(F 1 )P(F 2 F 1 )P(F 3 F 1 F 2 ) P(F k F 1 F k 1 ) = N N 1 N 2 1 N k + 1 = (N k)! N! For P(G E) see page 23. Thus, the final answer is: ( N ) (N k)! N k k N! i=0 ( 1)i /i!

12 of two events The formula for the probability of A B yields P(A B) = P(A) + P(B) P(A B). A simpler formula for P(A B) is possible if A and B are independent. Independent events arise (quite often but not always) in connection with independent experiments or independent repetitions of the same experiment. Thus there is no mechanism through which the outcome of one experiment will influence the outcome of the other. For example, two rolls of a die. For independent events, P(A B) = P(A)P(B). This also serves as the definition of independent events.

13 Example Toss a coin twice. To find the probability of two heads (Hs), we can argue that, since the two tosses are independent, P(H in toss 1 H in toss 2) = = 1 4 Alternatively, we can say that since P(H in toss 1 H in toss 2) = 1 4 = P(H in toss 1)P(H in toss 1), the events H in toss 1 and H in toss 2 are independent

14 Example 1. Select a card at random from a deck of 52 cards. Let E = {the card is an ace}, and F = {the card is a spade}. Are E and F independent? 2. Roll a die twice. Let E = {the sum of the two outcomes is 6}, and F = {the first roll results in 4}. Are E and F independent? 3. Same as above except that E = {the sum of two outcomes is 7}.

15 Proposition If E and F are independent, so are E and F c. Proof: P(E F c ) = P(E) P(E F ) = P(E)(1 P(F )) = P(E)P(F c ).

16 of Multiple Events Definition The events A 1,..., A n are mutually independent if P(A i1 A i2... A ik ) = P(A i1 )P(A i2 )... P(A ik ) for any sub-collection A i1,..., A ik of k events chosen from A 1,..., A n Example Consider rolling a die and define the events A = {1, 2, 3}, B = {3, 4, 5}, C = {1, 2, 3, 4}. Verify that P(A B C) = P(A)P(B)P(C), but that A, B, C are not mutually independent. Solution: First, A B C = {3}, so P(A B C) = 1 6 = 1 2 Next, A B = {3}, so P(A B) =

17 Example Roll a die twice. Let E = {the sum of the two outcomes is 7}, F = {the first roll results in 4}, and G = {the second roll results in 3}. Are E, F and G independent? ANSWER: E, F and G pairwise independent but So they are not independent. P(E F G) = 1.

18 Example 1. A system of n components connected in series fails if one component fails. If component i fails with a probability p i, independently of the others, what is the probability that the system fails? 2. What if the n components of the system are connected in parallel?

19 Example Two dice are rolled and the sum of the two outcomes is recorded. What is the probability that 5 happens before 7? Solution: If E n = {no 5 or 7 appear on the first n 1 rolls and a 5 appears on the nth}, then P( i=1e n ) = = P(E n ) i=1 i=1 ( )n = 2 5

20 Example There are n types of coupons, and each new one is independently of type i with probability p i. If each of k shoppers collects a coupon let A i = {at least one type i coupon has been collected}. Find P(A i A j ). Solution: Use independence and P(A i A j ) = P(A i ) + P(A j ) P(A i A j )

21 Let the events A 1, A 2..., A k be disjoint and make up the entire sample space, and let B denote an event whose probability we want to calculate, as in the figure B A A A A If we know P(B A j ) and P(A j ) for all j = 1, 2,..., k, the Law of Total Probability gives [see (3.4), p.73 and (3.1), p.65 in Ross] P(B) = P(A 1 )P(B A 1 ) + + P(A k )P(B A k )

22 : EXAMPLES 1. 40% of bean seeds come from supplier A and 60% come from supplier B. Seeds from supplier A have 50% germination rate while those from supplier B have a 75% rate. What is the probability that a randomly selected seed will germinate? ANSWER: P(G) = P(A)P(G A) + P(B)P(G B) = = Three players are dealt a card in succession. What is the probability that the 2nd gets a king? ANSWER: 4 52 Why?

23 Example Two dice are rolled and the sum of the two outcomes is recorded. What is the probability that 5 happens before 7? Solution: Let B be the desired event, A 1 = {first roll results in 5}, A 2 = {first roll results in 7}, A 3 = {first roll results in neither 5 not 7}. Then P(B) = P(B A 1 )P(A 1 ) + P(B A 2 )P(A 2 ) + P(B A 3 )P(A 3 ) = P(A 1 ) P(B)P(A 3 ).

24 Example (The Monte Hall Problem) In the game Let s Make a Deal, the host asks a participant to choose one of the doors A, B or C. Behind one of the doors is a big prize. The participant selects door A. The host then opens door B showing that the big prize is not behind it. The host asks the participant to either 1. stick with his/her original choice, or 2. select the other of the remaining two closed doors. Find the probability that the participant will win the big prize for each of the strategies a) and b).

25 Solution of the Monte Hall Problem Let A i = {prize is behind door i}, i = 1, 2, 3, and set B 1 = {player wins when choosing door 1 and does not change}, B 2 = {player wins when choosing door 1 and then changes}. Then, P(B 1 ) = P(B 2 ) = 3 P(B 1 A i )P(A i ) = = 1 3 i=1 3 P(B 2 A i )P(A i ) = = 2 3. i=1

26 Consider events B and A 1,..., A k as in the Law of Total Probability. Now, however, we ask a different question: Given that B has occurred, what is the probability that a particular A j has occurred? The answer is provided by the Bayes theorem: P(A j B) = P(A j B) P(B) = P(A j )P(B A j ) k j=1 P(A i)p(b A i )

27 : EXAMPLES 1. 40% of bean seeds come from supplier A and 60% come from supplier B. Seeds from supplier A have 50% germination rate while those from supplier B have a 75% rate. Given that a randomly selected seed germinated, what is the probability that it came from supplier A? ANSWER: P(A G) = P(A)P(G A) P(A)P(G A) + P(B)P(G B) = Given that the 2nd player got an ace, what is the probability that the 1st got an ace? ANSWER: 3 51 (Why?)

28 EXAMPLE: Law of Total Prob. and Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator, whereas 10% of the aircraft not discovered have such a locator. Suppose a light aircraft has disappeared. 1. What is the probability that it has an emergency locator and it will not be discovered? Answer: = What is the probability that it has an emergency locator? Answer: = If it has an emergency locator, what is the probability that it will not be discovered? Answer: 0.03/0.45

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