Tree Diagrams

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Questions 1-4

129

Question 1.

In the tree diagram below, what is the probability of B if event A does not happen?

0.45

0.79

0.75

0.55

Correct. If event A does not happen, we are on the lower set of branches. We can read that the probability of B happening in that case is 0.25.

Incorrect. If event A does not happen, we are on the lower set of branches. We can read that the probability of B happening in that case is 0.25.

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Question 2.

In the tree diagram below, what is the probability that event B does not happen if event A occurs?

0.45

0.25

0.79

0.25

Correct. Given that A occurs, we are on the top set of branches. In that case, we read P(B) = 0.55, so P(not B) = 1 – 0.55.

Incorrect. Given that A occurs, we are on the top set of branches. In that case, we read P(B) = 0.55, so P(not B) = 1 – 0.55.

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Question 3.

A fast food restaurant sells burgers, chicken sandwiches, and entree salads. If a customer buys a burger, the probability they buy french fries is 0.80. Fifty percent of customers who purchase a chicken sandwich also buy fries, but only 10% of those buying an entree salad buy fries. Of all customers, 65% buy a burger, 20% a chicken sandwich, the rest buy an entree salad.

In this situation as diagrammed, the 0.80 represents

P(Fries and Burger)

P(Fries given a Burger)

P(Fries and no Burger)

Correct. The 0.80 occurs on the second set of branches, so it is a conditional probability. The first event on that branch is buying a burger.

Incorrect. The 0.80 occurs on the second set of branches, so it is a conditional probability. The first event on that branch is buying a burger.

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Question 4.

A fast food restaurant sells burgers, chicken sandwiches, and entree salads. If a customer buys a burger, the probability they buy french fries is 0.80. Fifty percent of customers who purchase a chicken sandwich also buy fries, but only 10% of those buying an entree salad buy fries. Of all customers, 65% buy a burger, 20% a chicken sandwich, the rest buy an entree salad.

What is the probability a customer does not get fries if they get a chicken sandwich?

Correct. Each set of branches must add to 1. Because the probability of fries if a chicken sandwich is purchased is 0.50, the probability of not getting fries with the chicken sandwich is 0.50.

Incorrect. Each set of branches must add to 1. Because the probability of fries if a chicken sandwich is purchased is 0.50, the probability of not getting fries with the chicken sandwich is 0.50.

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Questions 5-8

189

Question 5.

In the tree diagram above, what is the probability events A and B both happen?

Correct. This is a joint probability; we find those by multiplying out the branches of the tree. Here, P(A and B) = 0.75*0.55.

Incorrect. This is a joint probability; we find those by multiplying out the branches of the tree. Here, P(A and B) = 0.75*0.55.

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Question 6.

In the diagram above, what is the probability of B, given A has happened?

0.55

0.44

0.50

0.80

Correct. We must have P(A)*P(B | A) = 0.44. Because we were told P(not A) = 0.45, we must have P(A) = 0.55. So, we have 0.55* P(B | A) = 0.44. Solve the equation for P(B | A).

Incorrect. We must have P(A)*P(B | A) = 0.44. Because we were told P(not A) = 0.45, we must have P(A) = 0.55. So, we have 0.55* P(B | A) = 0.44. Solve the equation for P(B | A).

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Question 7.

A fast food restaurant sells burgers, chicken sandwiches, and entree salads. If a customer buys a burger, the probability they buy french fries is 0.80. Fifty percent of customers who purchase a chicken sandwich also buy fries, but only 10% of those buying an entree salad buy fries. Of all customers, 65% buy a burger, 20% a chicken sandwich, the rest buy an entree salad. What proportion of customers get a burger and fries?

Correct. We want P(Burger and Fries). To find that probability, we multiply along that branch of the tree and have 0.65*0.80.

Incorrect. We want P(Burger and Fries). To find that probability, we multiply along that branch of the tree and have 0.65*0.80.

Try again.

Question 8.

A fast food restaurant sells burgers, chicken sandwiches, and entree salads. If a customer buys a burger, the probability they buy french fries is 0.80. Fifty percent of customers who purchase a chicken sandwich also buy fries, but only 10% of those buying an entree salad buy fries. Of all customers, 65% buy a burger, 20% a chicken sandwich, the rest buy an entree salad. What proportion of customers get a salad and no fries?

0.01

0.10

0.015

0.135

Correct. Because each set of branches must have probabilities adding to 1, we find P(Salad) = 1 − 0.65 − 0.20 = 0.15. Similarly, because P(Fries | Salad) = 0.10, we must have P(no Fries | Salad) = 0.90. Multiply 0.15*0.90 to find the probability of a salad without fries (the healthy meal).

Incorrect. Because each set of branches must have probabilities adding to 1, we find P(Salad) = 1 − 0.65 − 0.20 = 0.15. Similarly, because P(Fries | Salad) = 0.10, we must have P(no Fries | Salad) = 0.90. Multiply 0.15*0.90 to find the probability of a salad without fries (the healthy meal).

Try again.

Questions 9-10

409

Question 9.

In the tree diagram above, what is the probability event B happens?

0.67

1.34

0.61

0.55

Correct. To find the probability event B happens, we need to multiply out the two branches that end with B and add those together. We have 0.75*0.55 + (1 – 0.75)*0.79.

Incorrect. To find the probability event B happens, we need to multiply out the two branches that end with B and add those together. We have 0.75*0.55 + (1 – 0.75)*0.79.

Try again.

Question 10.

A fast food restaurant sells burgers, chicken sandwiches, and entree salads. If a customer buys a burger, the probability they buy french fries is 0.80. Fifty percent of customers who purchase a chicken sandwich also buy fries, but only 10% of those buying an entree salad buy fries. Of all customers, 65% buy a burger, 20% a chicken sandwich, the rest buy an entree salad. What proportion of customers do not get french fries?

0.400

1.60

0.635

0.365

Correct. To find the probability a customer gets french fries, we multiply out the branches that result in no french fries and add those results together. We have 0.65*(1 – 0.80) + 0.20*(1 – 0.50) + (1 – 0.65 – 0.20)*(1 – 0.10).

Incorrect. To find the probability a customer gets french fries, we multiply out the branches that result in no french fries and add those results together. We have 0.65*(1 – 0.80) + 0.20*(1 – 0.50) + (1 – 0.65 – 0.20)*(1 – 0.10).

Try again.

StatTutor Lesson - Tree Diagrams

Questions 1-4

Question 1.

Question 2.

Question 3.

Question 4.

Questions 5-8

Question 5.

Question 6.

Question 7.

Question 8.

Questions 9-10

Question 9.

Question 10.