7/20

The following table shows the number of people that like a particular fast food restaurant.

1) What is the probability that a person likes Wendy’s?

2) What is the probability that a person likes McDonald’s or Burger King?

3. What is the probability that a randomly chosen person is female or likes McDonald’s?

65/100 = 13/20

McDonald’s Burger King Wendy’s

Male 20 15 10

Female 20 10 25

3/4

Math I

UNIT QUESTION: How do you use probability to make plans and predict for the future?Standard: MM1D1-3

Today’s Question:When do I add or multiply when solving compound probabilities?Standard: MM1D2.a,b.

ProbabilityProbability

Conditional Probability Conditional Probability and Independent vs. and Independent vs. Dependent eventsDependent events

Conditional ProbabilityConditional ProbabilityA A conditional probability conditional probability is the probability of an is the probability of an event occurring, given that another event has event occurring, given that another event has already occurred. The conditional probability of already occurred. The conditional probability of event B occurring, given that event A has event B occurring, given that event A has occurred, is denoted by P(B/A) and is read as occurred, is denoted by P(B/A) and is read as “probability of B, given A.”“probability of B, given A.”

This is an “and” question, and solved by This is an “and” question, and solved by multiplicationmultiplication

Conditional ProbabilityConditional Probability Two cards are selected in sequence from a Two cards are selected in sequence from a

standard deck. Find the probability that the standard deck. Find the probability that the second card is a queen, given that the first card second card is a queen, given that the first card is a king, and we did not replace the king.is a king, and we did not replace the king.

Because the first card is a king and is not Because the first card is a king and is not replaced, the remaining deck has 51 cards, 4 of replaced, the remaining deck has 51 cards, 4 of which are queens, so P(B/A) = 4/51 which are queens, so P(B/A) = 4/51 0.0078 0.0078

Conditional ProbabilityConditional ProbabilityGene Present

Gene Not Present

Total

High IQ 33 19 52Normal IQ 39 11 50Total 72 30 102

The above table shows the results of a study in The above table shows the results of a study in which researchers examined a child’s IQ and which researchers examined a child’s IQ and the presence of a specific gene in the child. the presence of a specific gene in the child. Find the probability that a child has a high IQ, Find the probability that a child has a high IQ, given that the child has the gene.given that the child has the gene.

Conditional ProbabilityConditional ProbabilityGene Present

Gene Not Present

Total

High IQ 33 19 52Normal IQ 39 11 50Total 72 30 102

There are 72 children who have the gene so There are 72 children who have the gene so the sample space consists of these 72 children. the sample space consists of these 72 children. Of these, 33 have a high IQ, so Of these, 33 have a high IQ, so

P(B/A) = 33/72 P(B/A) = 33/72 0.458 0.458

Conditional ProbabilityConditional ProbabilityGene Present

Gene Not Present

Total

High IQ 33 19 52Normal IQ 39 11 50Total 72 30 102

Find the probability that a child does not have Find the probability that a child does not have the gene.the gene.

P(child does not have the gene) = 30/102P(child does not have the gene) = 30/102 Find the probability that a child does not have Find the probability that a child does not have

the gene, given that the child has a normal IQthe gene, given that the child has a normal IQ P(B/A) = 11/50P(B/A) = 11/50

What is the probability of the blood being What is the probability of the blood being type B given it is positive?type B given it is positive?

37/34437/344 What is the probability of the blood being What is the probability of the blood being

type RH Positive, given it is B or AB?type RH Positive, given it is B or AB? (37 + 12)/(45 + 16) = 49/61(37 + 12)/(45 + 16) = 49/61

Blood TypeO A B AB Tota

lRH Factor

Positive 156 139 37 12 344Negative 28 25 8 4 65Total 184 164 45 16 409

Independent EventsIndependent Events Two events A and B, are Two events A and B, are independentindependent if the fact that A if the fact that A

occurs does not affect the probability of B occurring.occurs does not affect the probability of B occurring. Then P(B/A) = P(B)Then P(B/A) = P(B) Examples - Landing on heads from two different Examples - Landing on heads from two different

coins, rolling a 4 on a die, then rolling a 3 on a second coins, rolling a 4 on a die, then rolling a 3 on a second roll of the die.roll of the die.

Probability of A and B occurringProbability of A and B occurring:: P(A and B)=P(A)*P(B) P(A and B)=P(A)*P(B)

ProbabilityProbability NOTE: NOTE: You add something to get the probability of You add something to get the probability of

something something OROR something something You multiply something to get the probability of You multiply something to get the probability of

something something ANDAND something. something.

Experiment 1Experiment 1 A coin is tossed and a 6-sided die is rolled. A coin is tossed and a 6-sided die is rolled.

Find the probability of landing on the head Find the probability of landing on the head side of the coin and rolling a 3 on the die.side of the coin and rolling a 3 on the die.

P (head)=1/2P (head)=1/2P(3)=1/6P(3)=1/6P (head and 3)=P (head)*P(3)P (head and 3)=P (head)*P(3) =1/2 * 1/6=1/2 * 1/6 = 1/12= 1/12

Experiment 2Experiment 2 A card is chosen at random from a deck of 52 A card is chosen at random from a deck of 52

cards. It is then replaced and a second card is cards. It is then replaced and a second card is chosen. What is the probability of choosing a chosen. What is the probability of choosing a jack and an eight?jack and an eight?

P (jack)= 4/52P (jack)= 4/52P (8)= 4/52P (8)= 4/52P (jack and 8)= 4/52 * 4/52P (jack and 8)= 4/52 * 4/52 = 1/169= 1/169

Experiment 3Experiment 3 A jar contains three red, five green, two blue A jar contains three red, five green, two blue

and six yellow marbles. A marble is chosen at and six yellow marbles. A marble is chosen at random from the jar. After replacing it, a random from the jar. After replacing it, a second marble is chosen. What is the second marble is chosen. What is the probability of choosing a green and a yellow probability of choosing a green and a yellow marble?marble?

P (green) = 5/16P (green) = 5/16P (yellow) = 6/16P (yellow) = 6/16P (green and yellow) = P (green) x P (yellow)P (green and yellow) = P (green) x P (yellow)

= 15 / 128= 15 / 128

Experiment 4Experiment 4 A school survey found that 9 out of 10 A school survey found that 9 out of 10

students like pizza. If three students are chosen students like pizza. If three students are chosen at random with replacement, what is the at random with replacement, what is the probability that all three students like pizza?probability that all three students like pizza?

P (student 1 likes pizza) = 9/10P (student 1 likes pizza) = 9/10P (student 2 likes pizza) = 9/10P (student 2 likes pizza) = 9/10P (student 3 likes pizza) = 9/10P (student 3 likes pizza) = 9/10P (student 1 and student 2 and student 3 like P (student 1 and student 2 and student 3 like pizza) = 9/10 x 9/10 x 9/10 = 729/1000pizza) = 9/10 x 9/10 x 9/10 = 729/1000

Dependent EventsDependent Events Two events A and B, are Two events A and B, are dependentdependent if the fact that if the fact that

A occurs affects the probability of B occurring.A occurs affects the probability of B occurring. Examples- Picking a blue marble and then Examples- Picking a blue marble and then

picking another blue marble if I don’t replace the picking another blue marble if I don’t replace the first one.first one.

Probability of A and B occurringProbability of A and B occurring:: P(A and B)=P(A)*P(B/A) P(A and B)=P(A)*P(B/A)

Experiment 1Experiment 1 A jar contains three red, five green, two blue A jar contains three red, five green, two blue

and six yellow marbles. A marble is chosen at and six yellow marbles. A marble is chosen at random from the jar. A second marble is random from the jar. A second marble is chosen chosen withoutwithout replacing the first one. What is replacing the first one. What is the probability of choosing a green and a the probability of choosing a green and a yellow marble?yellow marble?

P (green) = 5/16P (green) = 5/16P (yellow given green) = 6/15P (yellow given green) = 6/15P (green and then yellow) = P (green) x P (yellow)P (green and then yellow) = P (green) x P (yellow)

= 1/8= 1/8

Experiment 2Experiment 2 An aquarium contains 6 male goldfish and 4 An aquarium contains 6 male goldfish and 4

female goldfish. You randomly select a fish female goldfish. You randomly select a fish from the tank, from the tank, do notdo not replace it, and then replace it, and then randomly select a second fish. What is the randomly select a second fish. What is the probability that both fish are male?probability that both fish are male?

P (male) = 6/10P (male) = 6/10P (male given 1P (male given 1stst male) = 5/9 male) = 5/9P (male and then, male) = 1/3P (male and then, male) = 1/3

Experiment 3Experiment 3 A random sample of parts coming off a A random sample of parts coming off a

machine is done by an inspector. He found machine is done by an inspector. He found that 5 out of 100 parts are bad on average. If that 5 out of 100 parts are bad on average. If he were to do a new sample, what is the he were to do a new sample, what is the probability that he picks a bad part and then, probability that he picks a bad part and then, picks another bad part if he picks another bad part if he doesn’tdoesn’t replace the replace the first?first?

P (bad) = 5/100P (bad) = 5/100P (bad given 1P (bad given 1stst bad) = 4/99 bad) = 4/99P (bad and then, bad) = 1/495P (bad and then, bad) = 1/495

Class WorkClass Work Pg 353, # 5 – 8 all andPg 353, # 5 – 8 all and Handout: # 5-40 and 5-73 through 5-88Handout: # 5-40 and 5-73 through 5-88