Download Math 104 ACTIVITY 7: Conditional probability, the multiplication rule

Math 104
ACTIVITY 7: Conditional probability, the multiplication rule, and stochastic processes
Why
Conditional probability deals with the relation between events, measuring how the probability of an event
changes when other event s are known to occur. The multiplication rule is an important tool for understanding
events in multi-step experiments. The probability models for multi-step experiments are often called stochastic
processes.
LEARNING OBJECTIVES
1. Be able to work effectively as a team, using the team roles
2. Be able to find and interpret conditional probabilities in different situations.
3. Be able to use the multiplication rule to find joint (“and”) probabilities.
4. Be able to use tree diagrams to find and understand probabilities in multi-step situations.
CITERIA
1. Success in working as a team and in fulfilling the team roles.
2. Success in involving all members of the team in the conversation.
3. Success in completing the exercises
RESOURCES
1. The course syllabus
2. The team role desk markers (handed out in class for use during the semester)
3. Your text - Sections 3.2 - 3.3
4. 50 minutes
PLAN
1. Select roles, if you have not already done so, and decide how you will carry out steps 2 and 3.
2. Work through the exercises given here - be sure everyone understands all results & procedures.
3. Assess the team’s work and roles performances and prepare the Reflector’s and Recorder’s reports including team
grade.
DISCUSSION
The conditional probability for an event A given B is the revised probability for A in the situation
where B is known to occur. The computational formula is
P r[A ∩ B]
P r[A|B] =
P r[B]
From this we also get
The multiplication rule for probability for calculating joint probability
P r[A ∩ B] = P r[A] × P r[B|A] = P r[B] × P r[A|B]
(Notice that P r[A ∩ B] is not equal to P r[A] × P r[B] unless P r[B|A] is equal to P r[A]—in this case A
and B are independent)
This is particularly useful when a probability experiment involves several steps and the probabilities at each
step depend on what has already happened (conditional probabilities). We can often represent the sample
space and the probabilities for various outcomes by bringing back the idea of a tree diagram. As before, we
represent each step in the process by branching the tree—but now we add a label to each branch edge showing
the conditional probability for the result. To get the probability of an outcome—represented by a branch from
the start to the end of the tree—we multiply the (conditional) probabilities along the branch. To get
the probability of an event, we add the probabilities for the outcomes in the event.
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MODELS
1. During the winter in East Overshoe, the probability of snow on a randomly chosen day is .45, the probability of
below 0 temperature is .35 and the probability of both is .10.
and below 0]
On a day when the temperature is below 0, the probability of snow is P r[snow|below 0] = P r[snow
=
P r[below 0]
2
.10
.35 = 7 ≈ .29
and below 0
= .10
On a day when it snows, the probability of a temperature below 0 is P r[below 0|snow] = P R[snow
P r[snow]
.45 =
2
≈
.22
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2. The probability that Susan will attend her Math class on a randomly chosen day is .75; if she attends Math
class, the probability she will also attend her English class is .80. The probability that she will attend both is
P r[Math ∩ English] = P r[Math] × P r[English| Math] = .75 × .80 = .60
Now we can also say that the probability that she attends at least one of these classes is P r[Math ∪ English] =
P r[Math] + P r[English] − P r[Math and English] = .75 + .80 − .60 = .95
3. Back at the College of Knowledge, 30% of the students are Freshmen. If a student is a Freshman, there is a 75%
chance she eats in the dining hall. If a student is not a Freshman, there is a 60% chance she eats in the dining hall.
We are interested in some probabilities when we pick a student at random.
(a) The probability of getting a Freshman who eats in the dining hall
(b) The probability of getting a Freshman who does not eat in the dining hall
(c) the probability of getting a non-Freshman who eats in the dining hall
(d) The probability of getting a student who eats in the dining hall
(e) the probability of getting a Freshman, if we pick a student who eats in the dining hall
We can represent the probability situation with a tree diagram - first branching: Freshman or not: We know
P r[Freshman] = .30 so P r[not Freshman] = .70
Second branching: dining hall, or not; We know P r[dining hall|Freshman] = .75 and P r[dining hall|not Freshman] =
.60. So our tree is:
The numbers on the branches are the conditional probabilities for the next step. To obtain the probability of an
outcome, we multiply along the path leading to that outcome.
(NOTE: We could draw a similar diagram with “dining hall (Yes/No)” first—but we do not know the numbers
needed for labeling that tree.)
Then we can answer the questions given:
(a) P r[Freshman and dining hall] = .225 – note this is a joint probability, not conditional (we are looking for both
conditions – we don’t know that either one has occurred)
(b) P r[Freshman and not dining hall] = .075
(c) P r[Not Freshman, dining hall] = .42
(d) P r[dining hall] = P r[Freshman and dining hall] + P r[not Freshman, dining hall] = .225 + .42 = .645
and dining hall]
(e) P r[Freshman|dining hall] = P r[Freshman
= .225
P r[dining hall]
.645 ≈ .349 – Note this is a conditional probability we already know we have a student who eats in the dining hall.
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EXERCISES
1. E and F are events in a sample space S, with P r[E] = .45, P r[F ] = .60, P r[E ∪ F ] = .85 Answer the following (you
probably will want a Venn diagram—why is it not likely that a tree diagram will be useful here?):
(a) Give P r[E ∩ F ]
(b) What tells you that E and F are not independent?
(c) Give P r[E|F ]
(d) Give P r[E 0 |F ]
(e) Give P r[F |E]
2. United Widget Company makes 40% of all widgets sold in the U.S. Of the widgets they make, 3% are defective. If
a widget sold in the U. S. is selected at random, what is the probability it is from United Widget and is defective?
3. A box contains 5 red balls, 3 white balls and 2 green balls. Two balls will be taken at random, without replacement,
and the colors noted (in order).
(a) If the first ball is red, what is the (conditional) probability that the second ball is also red? (You don’t need a
formula for this)
(b) Draw the tree diagram for this experiment – show the probabilities on the branches (as in the model).
(c) What is the probability of getting two red balls?
(d) What is the probability of getting exactly one red ball?
4. Two friends A and B play a checkers match which will end when one player has won three games (then that player
is the winner) or when four games have been played (if each has won two games, the match is a tie). Player A is
slightly better; the probability that she will win the first game is .6. When a player wins a game, her probability
of winning increases by .1 (if A wins the first game, the probabilities for the second are .6 for A and .3 for B; if B
wins the first game the probabilities for the second are .5 for A and .5 for B; if A wins the first two, her probability
of winning the third goes up to .8, etc.)
(a) Draw the tree and calculate the probabilities for for each of the outcomes.
(b) What is the probability that A wins the match in three games?
(c) What is the probability that A wins the match at the fourth game? (there are several outcomes in this event)
(d) What is the probability that B wins the match?
(e) What is the probability that the match ends in a tie?
(f) If B wins the first game, what is the probability she will win the match?
READING ASSIGNMENT (in preparation for next class meeting)
Read Section 3.4 in the text (Don’t worry about the formula or the name ”Bayes”, do pay attention to the
trees and the calculations - all based on “multiply along, add across” and the definition/formula for conditional
probabilty)
SKILL EXERCISES:(hand in - individually - at next class meeting)
p.114 #20–21, 23–25, 27-28,
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