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Sections 4.4 and 4.5: The Multiplication Rule
and Conditional Probability
Since the size of a sample space grows so quickly we want to continue our search for rules of that allow us to compute the probabilities of complex events. When thinking about what happens with combinations of outcomes, things are simpliÖed if the individual trials are independent.
DeÖnition 1 Two events are independent if the outcome of one event doesnít ináuence or change the likelihood of the outcome of the other event.
Example 1 Coin áips are independent. The coin does not remember its se- quence of áips; the chance of heads or tails is always constant at p = 12.
Example 2 From ajc.com on 1/6/2017, "As the metro area faces another dire winter forecast, itís hard to forget that only three years ago we let 2.6 inches of snow knock us on our collective backside, turning Atlanta into national laughingstock. We called it ìSnowJam ë14.î Not to be confused with ìSnow Jam ë82î where nearly the same thing happened, or ìSnowpocalypse ë11î which had been so recent that some leaders Ögured the region was statistically safe from another snow debacle for at least a decade."
Example 3 The Phillies chance of winning the World Series in 2017 and the health of Phillies pitchers are dependent events. If any pitcher su§ ers a serious injury and is out for the season, the Phillies chance of winning the World Series goes down.
Example 4 The Phillies chance of winning the World Series in 2017 and the health of pitchers of their opponents are dependent events. If any pitcher su§ ers a serious injury and is out for the season, their teamís chance of winning goes down and the Phillies chance of winning the World Series goes up.
Example 5 The Phillies chance of winning the World Series in 2017 and the health of Dr. DeMaioís pitching arm are independent events. No matter what happens to Dr. DeMaio, the Phillies chances of winning the World Series are completely una§ected.
Theorem 1 Multiplication Rule: For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events.
P (A and B) = P (A) P (B)
Example 6 Approximately 85% of all human beings are right-handed. What is the probability that three randomly selected people are all right-handed? p = : 85 : 85 :85 = : 853 = 0:614 13.
Example 7 Ignoring ambidextrous people, What is the probability that two ran- domly selected people are all left-handed? p = : 152 = 0:022 5.
Exercise 1 Shaquille OíNealís lifetime free throw percentage is .527. Shaq is fouled on a three-point shot. What is the probability that he makes all three free throws?
What is the probability that he makes none of the free throws?
Exercise 2 A box contains 3 white balls, 4 red balls and 5 black balls. A ball is picked, its color recorded and returned to the box. Another ball is then selected and its color recorded.
Remark 2 Since we put the 1st ball back into the box before selecting another, we are making selections with replacement. Doing so makes subsequent selections independent events. Find the probability that 2 black balls are selected.
Find the probability that 2 balls of the same color are selected.
Example 8 A box contains 3 white balls, 4 red balls and 5 black balls. Four balls are picked with replacement. Find the probability no red balls are selected.
Find the probability that the fourth ball selected is the Örst occurrence of the color white?
A requirement of the multiplication rule is that events are independent. Nat- urally, this will not always be the case. In order to compute the probability of A and B when they are not independent events we rely on conditional probabil- ities. When we want the probability of an event from a conditional distribution, we write P (BjA) and say ìthe probability of B given A.î A probability that takes into account a given condition is called a conditional probability.
Theorem 3 General Multiplication Rule: For any two events A and B, the probability that both A and B occur is the.
P (A and B) = P (A) P (BjA)
Exercise 4 A study at a local bar found people of various ages playing games.
21 - 29 30 - 39 40 - 49 50 and older Total
Darts 4 12 15 6 37
Pool 8 17 16 11 52
Karaoke 17 5 0 1 23
Total 29 34 31 18 112
Find the probability that a randomly selected person...
- Plays darts.
- Is 21-29.
- Is 21-29 given that they are playing darts.
- Is 21-29 given that they are singing karaoke.
- Is singing karaoke given that they are 21-29.
- Is 30-39 and playing pool.
- Is playing pool given that they are 30-39.
A pair of dice is thrown one at a time. Let A be the event that the sum of 9 is rolled. Let B be the event that the Örst die thrown is a 2. Let C be the event that the Örst die thrown is a 5. Let D be the event that the sum of 7 is rolled.
- What is the probability the sum of the dice is 9?
- What is the probability the sum of the dice is 9, given that the Örst die rolled is 2?
- What is the probability the sum of the dice is 9, given that the Örst die rolled is 5?
- Are events A and B independent? Are events A and C independent?
- What is the probability the sum of the dice is 7?
- What is the probability the sum of the dice is 7, given that the Örst die rolled is 2?
- What is the probability the sum of the dice is 7, given that the Örst die rolled is 5?
- Are events D and B independent? Are events D and C independent?
Exercise 5 A box contains 3 white balls, 4 red balls and 5 black balls. A ball is picked, its color recorded and set aside. Another ball is then selected and its color recorded.
Remark 4 In this case, we did not return the 1st ball back to the box before selecting another. We are now making selections without replacement. Do- ing so makes subsequent selections dependent or conditional events. Find the probability that 2 black balls are selected.
Find the probability that 2 balls of the same color are selected.
Find the probability that the second ball selected is the Örst occurrence of the color white?
Knowing that P (A and B) = P (A) P (BjA)
allows us to rearrange terms and see that
P (BjA) =
P (A and B) P (A)
Example 12 For residents of Atlanta, 35% are fans of the Braves, 45% are fans of the Falcons and 13% are fans of both the Braves and the Falcons.
- Given that a person is a Falcons fan, what is the probability that they are a Braves fan? P (BjF ) = P^ (B P^ and(F )^ F^ )= : :^1345 = 0:288 89:
- What is the probability that a person is neither a Braves fan nor a Falcons fan? 1 P (B or F ) = 1 (:35 + : 45 :13) = 0: 33.
- Given that a person is not a Braves fan, what is the probability that they are not a Falcons fan? P ( F j B) = P^ ( B and F ) P ( B) =^
: 33 : 65 = 0:507 69.
1 Exercises
- Kokoska 3rd edition Section 4.4: 4.130, 4.132, 4.133, 4.134 a, c, d, 4.135, 4.137, 4.138, 4.140-4.
- Kokoska 3rd edition Section 4.5: 4.159-4.163, 4.168-4.171, 4.175-4.181, 4.185, 4.193,
