{ } all possible outcomes of the procedure. There are 8 ways this procedure can happen.

Transcription

1 Probability with the 3-Kids Procedure Statistics Procedures and Events Definition A procedure is something that produces an outcome. When a procedure produces an outcome, it s called a trial or a run of the procedure. Example Here s a procedure: Have three children and record their genders in order of birth. Let s call this the 3-Kids procedure. In one trial of the 3-Kids procedure, there s an outcome of bgb (boy-girl-boy). In another trial you can get gbb. The set of all possible outcomes for a procedure is the procedure s sample space. The sample space for the 3-Kids procedure is bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg since this is the set of all possible outcomes of the procedure. There are ways this procedure can happen. An event is a subset of the sample space containing outcomes that are said to result in the event. An outcome that results in an event is a way the event happened. Here s an event that can happen as a result of the 3-Kids procedure. bgg, gbg, ggb. A = Having exactly two girls = How many ways can A happen? Three. And another event. B = The oldest is a girl = {,,, } gbb gbg ggb ggg. Does the outcome of gbg result in B happening? Certainly. A single outcome considered as an event is called a simple event. Let event C = Having all boys = { bbb }. Since there s only one way for this event to happen, we call it a simple event The complement of event A consists of all the ways that A doesn t happen. It's written as A and pronounced A complement. In our case, A = not getting exactly two girls = bbb, bbg, bgb, gbb, ggg and can happen 5 ways.

2 A compound event is an event that s made up of other events. (A and B) = all the ways that result in both A and B happening at the same time. (A or B) = all the ways that result in either A or B or both happening. The compound event (A and B) = Having exactly two girls gbg, ggb. There are only and the oldest is a girl (too) = ways this can happen. The compound event (A or B) = Having exactly two girls or bgg, gbb, gbg, ggb, ggg. having the oldest be a girl. = There s 5 ways this can happen. Probability P( A ) = The probability that the event A will occur (when running the procedure). In the 3-Kids example, P( A ) = The probability that there will be exactly two girls. To compute the (theoretical) probability of an event A happening: Count the number of ways A can happen, and divide by the total number of ways the procedure can happen. P A # of ways A can happen =. # of ways procedure can happen The # of ways the procedure can happen is the size of the sample space. By counting in the 3-Kids sample space, we know there are 3 ways event A can happen. The size of the sample space is, so 3 P( A ) =. Similarly, 4 P( B ) = = and P( C ) =. We also have the probability of compound events, 5 P( Aand B ) = = and P( Aor B ) =. 4 The probability of the complement of an event is one minus the probability of the event. PA = PA. Using the formula what s the probability of not having exactly two girls? 3 5 PA = PA = =.

3 Two events are disjoint if there is no way they can both happen at the same time (in the same run of the procedure). If A and B are Aand B disjoint, then the compound event cannot happen. So P( Aand B ) = 0. If two events are not disjoint, then we say they overlap. If A and B overlap, then P Aand B > 0. The 3-Kids events A and C are disjoint because there s no way to A = "have exactly two girls" and C = "all boys" at the same time. P Aand C = 0. So Events A and B overlap since there are two ways to A = "have exactly two girls" and B = "the oldest be a girl". The overlap is gbg, ggb and (A and B) = P( Aand B ) = =. 4 Addition Rule Overlapping Events The probability of the compound event (A or B) can be computed with ( or ) = + ( and ) P A B P A P B P A B Disjoint Events If A and B are disjoint, then so the Addition Rule becomes ( or ) = + P A B P A P B P Aand B = 0, Addition Rule Overlapping Events Events A and B overlap. We already know 5 P( Aor B ) = from counting in the sample space. But also using the formula we have ( or ) = + ( and ) P A B P A P B P A B = + = Disjoint Events Events A and C are disjoint. So, 3 P( Aor C) = P( A) + P( C) = + = Independent and Two events are independent if the probability of one event is not affected by the occurrence of the other. If two events are not independent, we say they are dependent. The event B="oldest being a girl" does not affect the chances of the D="youngest being a boy". So, B and D are independent events. The chances of B = the oldest being a girl improve once we know that A = having exactly two girls has happened. So, B and A are dependent events.

4 Conditional Probability To deal with dependent events, we use conditional probability. P( B A ) reads the probability of B given A and is the probability of B happening given that A has happened. If events A and B are dependent, then A P B This says that A happening has an effect on the probability of B happening. Conditional Probability We know that the probability of the oldest being a girl is P( B ) =. But computing P B A we count only within A = { bgg, gbg, ggb } and see that B can happen ways { gbg, ggb} out of the 3 ways that A can happen. So, P( B A ) =. Hence, 3 the chances of B happening do improve when you know A has already happened. A and B A P B are dependent events since If events A and B are independent, then A = P B This says the probability of B happening is the same whether A happens or not. How to Compute A, count the number of To compute ways that B can happen out of the ways that A can happen. Divide that by the number of ways A can happen. You could say that the event A becomes a new sample space out of which you compute the probability of B. D = the youngest is a boy and B = the oldest being a girl are independent events. So, we should find that P D B = P D. Since D = { bbb, bgb, gbb, ggb }, 4 P( D ) = = Let s show B has no affect on the chances of D by showing P( D B ) = as well. To P D B, count among the ways compute that B = { gbb, gbg, ggb, ggg} can happen to see that D can happen in ways gbb, ggb out of the possible 4, so P( D B ) = =. 4 Therefore, P( D B) P( D) are independent. = and the events

5 Multiplication Rule The probability of the compound event (A and B) can be computed with the formula ( and ) = P A B P A P B A If the events are independent, then A = P B and the Multiplication Rule simplifies to multiplying the separate probabilities ( and ) = P A B P A P B Multiplication Rule Using the multiplication rule to calculate P Aand B you d get, 3 P( Aand B) = P( A) P( B A) = = 3 4 With the independent events B and D, we d have P( Band D) = P( B) P( D) = = 4 This indicates there are only ways that (B and D) can happen. Those ways are (B and ggb, gbb D) = Conditional Probability - Formula The conditional probability P( B A ) may be computed with this formula P B A P B = ( and A) P( A) (This works because the # of ways B can happen out of the ways that A can happen is really the same as the # of ways A and B can happen together. ) Conditional Probability - Formula P B A and P D B P( B and A) 4 = = =. P( A) 3 3 P( Dand B) 4 = = = P( B)