Introductory Probability - PDF Free Download

Introductory Probability Conditional Probability: Independent Events and Intersections Dr. Nguyen nicholas.nguyen@uky.edu Department of Mathematics UK February 15, 2019

Agenda Independent Events and Intersections Two Events Examples Multiple Events Announcement: The fth homework is available. There is a quiz next Friday.

Independent Events Denition (4.1) Let E and F be two events. Then they are independent if either: P(E F ) = P(E) or P(F E) = P(F ). One or both events has probability zero.

Independent Events Denition (4.1) Let E and F be two events. Then they are independent if either: P(E F ) = P(E) or P(F E) = P(F ). One or both events has probability zero. Knowledge that E (or F ) occurred does not aect the probability of F (or E ).

Let E and F be two events. Then they are independent if either: P(E F ) = P(E) or P(F E) = P(F ). One or both events has probability zero. Knowledge that E (or F ) occurred does not aect the probability of F (or E ). Theorem (4.1) Two events E and F are independent if and only if P(E F ) = P(E)P(F ).

Four Statements Let E and F be events. The following four statements are either all true or all false: 1. The events E and F are independent. 2. P(E F ) = P(E) P(F ). 3. P(E F ) = P(E). 4. P(F E) = P(F ). To show whether E and F are independent or not, we need to compute P(E), P(F ), and at least one of P(E F ), P(E F ), and P(F E), and see whether the corresponding equality holds or not.

Sequences of 3 Tosses Overview Suppose we toss a fair coin 3 times. Let Ω be the 2 3 = 8 sequences of tosses. Let A 1 be the event The rst toss lands Heads. Let A 2 be the event There is 1 Head among the last two tosses. Let E be the event The rst toss lands Heads. Let F be the event There are 2 Heads total. We see if A 1 and A 2, and if E and F, are independent events.

Sequences of 3 Tosses 1a There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in A 1 are in yellow, those in A 2 are blue, those in both are green. P(A 1 ) = 4/8 = 1/2 P(A 2 ) = 4/8 = 1/2 Let's compute P(A 1 A2 ) and compare it to P(A 1 ) P(A 2 ): P(A 1 A 2 ) = 2/8 = 1/4 P(A 1 A 2 ) = 1 4 = 1 2 1 2 = P(A 1) P(A 2 ).

There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in A 1 are in yellow, those in A 2 are blue, those in both are green. P(A 1 ) = 4/8 = 1/2 P(A 2 ) = 4/8 = 1/2 Let's compute P(A 1 A2 ) and compare it to P(A 1 ) P(A 2 ): P(A 1 A 2 ) = 2/8 = 1/4 P(A 1 A 2 ) = 1 4 = 1 2 1 2 = P(A 1) P(A 2 ). Since P(A 1 A2 ) = P(A 1 ) P(A 2 ), A 1 and A 2 are independent.

Sequences of 3 Tosses 1b There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in A 1 are in yellow, those in A 2 are blue, those in both are green. Let's compute P(A 1 A 2 ) and compare it to P(A 1 ). We'll do this twice for practice. First, let's use our formula: P(A 1 A 2 ) = P(A 1 A 2 ) P(A 2 ) = 1/4 1/2 = 1 2.

There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in A 1 are in yellow, those in A 2 are blue, those in both are green. Let's compute P(A 1 A 2 ) and compare it to P(A 1 ). We'll do this twice for practice. First, let's use our formula: P(A 1 A 2 ) = P(A 1 A 2 ) P(A 2 ) = 1/4 1/2 = 1 2. Let's compute again using A 2 as the sample space: P(A 1 A 2 ) 2 outcomes with H rst and 1 H among last 2 tosses = 4 outcomes with 1 H among last 2 tosses = 1 2.

Let's compute P(A 1 A 2 ) and compare it to P(A 1 ). We'll do this twice for practice. First, let's use our formula: P(A 1 A 2 ) = P(A 1 A 2 ) P(A 2 ) = 1/4 1/2 = 1 2. Let's compute again using A 2 as the sample space: P(A 1 A 2 ) 2 outcomes with H rst and 1 H among last 2 tosses = 4 outcomes with 1 H among last 2 tosses We see that P(A 1 A 2 ) = 1 2 = P(A 1), so since P(A 1 A 2 ) = P(A 1 ), A 1 and A 2 are independent. = 1 2.

Sequences of 3 Tosses 1c There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in A 1 are in yellow, those in A 2 are blue, those in both are green. Let's compute P(A 2 A 1 ) and compare it to P(A 2 ). We'll do this twice for practice. First, let's use our formula: P(A 2 A 1 ) = P(A 1 A 2 ) P(A 1 ) = 1/4 1/2 = 1 2.

There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in A 1 are in yellow, those in A 2 are blue, those in both are green. Let's compute P(A 2 A 1 ) and compare it to P(A 2 ). We'll do this twice for practice. First, let's use our formula: P(A 2 A 1 ) = P(A 1 A 2 ) P(A 1 ) = 1/4 1/2 = 1 2. Let's compute again using A 1 as the sample space: P(A 2 A 1 ) 2 outcomes with H rst and 1 H among last 2 tosses = 4 outcomes with H rst = 1 2.

Let's compute P(A 2 A 1 ) and compare it to P(A 2 ). We'll do this twice for practice. First, let's use our formula: P(A 2 A 1 ) = P(A 1 A 2 ) P(A 1 ) = 1/4 1/2 = 1 2. Let's compute again using A 1 as the sample space: P(A 2 A 1 ) 2 outcomes with H rst and 1 H among last 2 tosses = 4 outcomes with H rst We see that P(A 2 A 1 ) = 1 2 = P(A 2), so since P(A 2 A 1 ) = P(A 2 ), A 1 and A 2 are independent. = 1 2.

Sequences of 3 Tosses 1 Intuition: Events A 1 (rst toss is heads) and A 2 (one head among the last two tosses) involve dierent sets of coin tosses, so knowledge about one event should not tell us anything about the other event.

Sequences of 3 Tosses 2a There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Let E be the event rst toss is heads and let F be the event there are exactly two heads. Those in E are yellow, those in F are blue, those in both are green. P(E) = 4/8 = 1/2 P(F ) = 3/8 Let's compute P(E F ) and compare it to P(E) P(F ): P(E F ) = 2/8 = 1/4 P(E F ) = 1 4 1 2 3 = P(E) P(F ). 8

There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Let E be the event rst toss is heads and let F be the event there are exactly two heads. Those in E are yellow, those in F are blue, those in both are green. P(E) = 4/8 = 1/2 P(F ) = 3/8 Let's compute P(E F ) and compare it to P(E) P(F ): P(E F ) = 2/8 = 1/4 P(E F ) = 1 4 1 2 3 = P(E) P(F ). 8 Since P(E F ) P(E) P(F ), E and F are not independent.

Sequences of 3 Tosses 2b There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in E are yellow, those in F are blue, those in both are green. Let's compute P(F E) and compare it to P(F ). We'll use E as the sample space: P(F E) 2 outcomes with H rst and 2 H's total = = 1 4 outcomes with H rst 2.

There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Those in E are yellow, those in F are blue, those in both are green. Let's compute P(F E) and compare it to P(F ). We'll use E as the sample space: P(F E) = 2 outcomes with H rst and 2 H's total 4 outcomes with H rst = 1 2. This is not equal to P(F ), which was 3/8. Hence E and F are not independent.

Sequences of 3 Tosses 2c There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Let's compute P(E F ) with F as the sample space: P(E F ) 2 outcomes with H rst and 2 H's total = = 2 3 outcomes with exactly 2 H's 3.

There are eight equally likely sequences of 3 tosses: HHH,HHT,HTH,THH, HTT,THT,TTH,TTT. Let's compute P(E F ) with F as the sample space: P(E F ) 2 outcomes with H rst and 2 H's total = = 2 3 outcomes with exactly 2 H's 3. This is not equal to P(E), which was 1/2. Hence E and F are not independent.

Independence with More Events Denition (4.2) Let {A 1,...,A n } be a set of n events. Then the set is mutually independent if for any subset {A i,...,a m } of 2, 3, 4,..., n events, P(A i... A m ) = P(A i )... P(A m ). We have to test intersections of any 2, 3, 4,..., n events at a time, not just 2 at a time!

Suits and Ranks Let us draw a card from a deck of 52 cards. P(E 1 ) = P(card has rank A) = 4 aces 52 cards = 1 13. P(E 2 ) = P(card is red) = P(E 3 ) = P(card is a heart) = 26 red cards = 1 52 cards 2. 13 hearts 52 cards = 1 4.

Suits and Ranks Let us draw a card from a deck of 52 cards. P(E 1 ) = P(card has rank A) = 4 aces 52 cards = 1 13. P(E 2 ) = P(card is red) = P(E 3 ) = P(card is a heart) = P(E 1 E 2 ) = P(card is red A) = 26 red cards = 1 52 cards 2. 13 hearts 52 cards = 1 4. 2: A or A = 1 52 cards 26.

Let us draw a card from a deck of 52 cards. P(E 1 ) = P(card has rank A) = 4 aces 52 cards = 1 13. P(E 2 ) = P(card is red) = P(E 3 ) = P(card is a heart) = P(E 1 E 2 ) = P(card is red A) = 26 red cards = 1 52 cards 2. 13 hearts 52 cards = 1 4. 2: A or A 52 cards P(E 1 E 3 ) = P(card is A ) = 1 A 52 cards = 1 52. P(E 2 E 3 ) = P(card is a heart) = 13 hearts 52 cards = 1 4. = 1 26.

P(E 1 ) = P(card has rank A) = 4 aces 52 cards = 1 13. P(E 2 ) = P(card is red) = P(E 3 ) = P(card is a heart) = P(E 1 E 2 ) = P(card is red A) = 26 red cards = 1 52 cards 2. 13 hearts 52 cards = 1 4. 2: A or A 52 cards P(E 1 E 3 ) = P(card is A ) = 1 A 52 cards = 1 52. P(E 2 E 3 ) = P(card is a heart) = 13 hearts 52 cards = 1 4. P(E 1 E 2 E 3 ) = P(card is A ) = 1 A 52 cards = 1 52. = 1 26.

Suits and Ranks E 1 and E 2 are independent: P(E 1 )P(E 2 ) = P(E 1 E 2 ) 1 13 1 2 = 1 26 E 1 and E 3 are independent: P(E 1 )P(E 3 ) = P(E 1 E 3 ) 1 13 1 4 = 1 52 E 2 and E 3 are not independent: P(E 2 )P(E 3 ) P(E 2 E 3 ) 1 2 1 4 1 4

E 1 and E 3 are independent: P(E 1 )P(E 3 ) = P(E 1 E 3 ) 1 13 1 4 = 1 52 E 2 and E 3 are not independent: P(E 2 )P(E 3 ) P(E 2 E 3 ) 1 2 1 4 1 4 E 1, E 2, and E 3 are not mutually independent for two reasons: P(E 2 )P(E 3 ) P(E 2 E 3 ), P(E 1 )P(E 2 )P(E 3 ) P(E 1 E 2 E 3 ) 1 13 1 4 1 2 1 52.

Suits and Ranks Intuition Knowledge of a card's color (suit) gives information about which suit (color) it could be. For example, A red card must be a heart or diamond. A heart must be a red card.

@Home: Reading Note: page numbers refer to printed version. Add 8 to get page numbers in a PDF reader. You should look at the urn example on pages 135-136. Please look at Examples 4-7 to 4.9 on pages 140-141. Try Exercise 7 on page 151 to see an example of three events where any pair of them are independent, but the three events are not mutually independent. Note that there are 4 equally likely sequences of 2 tosses: HH, HT, TH, TT: A = {HH, HT }, B = {HH, TH}, C = {HH, TT }.

Next Time Please read Section 4.1 (you can skip the historical remarks). Look at joint distributions. You may want to brush up on multivariable functions (functions with more than one input). The fth homework is due February 27.