Exercise 72 Page 508

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A snack cracker company conducted a taste test for three types of their crackers. They surveyed 250 people per age group from the table below. |l|c|c|c|c| Age & Cracker A & Cracker B & Cracker C [0.5em] Under 20 & 152 & 54 & 44 [0.5em] 20 to 39 & 107 & 85 & 58 [0.5em] 40 to 59 & 78 & 101 & 71 [0.5em] 60 and over & 34 & 68 & 148 [0.5em] We want to find the probability that a participant chose cracker A, or that they were under 20 years old. To do so, we will use the Addition Rule of Probability.

Addition Rule of Probability
For two events A and B, the probability that A or B occurs can be found using the following formula. P(AorB) = P(A) + P(B) - P(A and B)

First, let's notice that since there are 4 age groups and the company surveyed 250 people per age group, overall, the company surveyed 4 * 250 = 1000 people. Next, we will identify our events. In our case, Event A is that a person chose cracker A, and Event B is that a person was under 20. Event A : & Person chose cracker A Event B : & Person was under 20 Now, we will look again at our table and we will identify the elements that correspond to the occurrence of: Event A, Event B, and both Event A and B. Let's do it! |l|c|c|c|c| Age & Cracker A & Cracker B & Cracker C [0.5em] Under 20 & 152 & 54 & 44 [0.5em] 20 to 39 & 107 & 85 & 58 [0.5em] 40 to 59 & 78 & 101 & 71 [0.5em] 60 and over & 34 & 68 & 148 [0.5em] We will find the probabilities of: Event A, Event B, and both Event A and B. To do so, we will sum the numbers corresponding to these events, making sure to include the number of times both events happened in every calculation. In each case, we will divide the sum by 1000, the total number of surveyed people. Let's start with Event A. P(Event A) &= 152 + 107 + 78 + 34/1000 &=371/1000 &= 0.371 The probability that cracker A was chosen equals 0.371, or 37.1 %. Now, let's find the other two probabilities.

Event	Calculation	Result
Event A	152 + 107 + 78 + 34/1000	37.1 %
Event B	152 + 54 + 44/1000	25 %
Both Event A and B	152/1000	15.2 %

Finally, now that we have all the probabilities let's substitute them into the formula for the Addition Rule of Probability.

P(AorB) = P(A) + P(B) - P(A and B)

SubstituteValues

Substitute values

P(AorB) = 37.1 % + 25 % - 15.2 %

AddSubTerms

Add and subtract terms

P(AorB) = 46.9 %

The probability that cracker A was chosen or the person was under 20 years old equals 46.9 %.

b We want to find the probability that a participant did not choose cracker A and was over 20 years old. To do so, we will use the Complement Rule.

Complement Rule
If A is an event and A' is its complement, the following formula holds. P(A) + P(A') = 1

In our case, Event A is that a participant did not choose cracker A and was over 20 years old. The complement A' of that event is that a person chose cracker A or is under 20 years old. In Part A, we found that the probability of the complement is 46.9 %. Let's substitute this value into the formula to find P(A).

P(A) + P(A') = 1

Substitute

P(A')= 46.9 %

P(A) + 46.9 % = 1

WritePercent

Convert to percent

P(A) + 46.9 % = 100 %

SubEqn

LHS-46.9 %=RHS-46.9 %

P(A) = 53.1 %

The probability that a person did not choose cracker A and was over 20 years old equals 53.1 %.

c We want to find the probability that a participant was 20 years old or older using a complement to answer. To do so, let's recall the Complement Rule.

Complement Rule
If A is an event and A' is its complement, the following formula holds. P(A) + P(A') = 1

In our case, Event A is that a participant is 20 years old or older. The complement A' of this event is that a participant is under 20 years old. From Part A, we know that the probability that a person was under 20 years old is 25 %. Let's substitute this value into the formula to find P(A).

P(A) + P(A') = 1

Substitute

P(A')= 25 %

P(A) + 25 % = 1

WritePercent

Convert to percent

P(A) + 25 % = 100 %

SubEqn

LHS-25 %=RHS-25 %

P(A) = 75 %

The probability that a participant was 20 years old or older equals 75 %.

d A randomly-selected person claims that he is 15 years old. We want to estimate the probability that he chose cracker A. To do so, let's recall the results of the survey, focusing on the entry for participants that are under 20 years old and chose cracker A.

|l|c|c|c|c| Age & Cracker A & Cracker B & Cracker C [0.5em] Under 20 & 152 & 54 & 44 [0.5em] 20 to 39 & 107 & 85 & 58 [0.5em] 40 to 59 & 78 & 101 & 71 [0.5em] 60 and over & 34 & 68 & 148 [0.5em] There are 152 surveyed people who were under 20 years old and chose cracker A. We also know that within this age group 250 people were surveyed. The desired probability is the ratio between the number of people under 20 years old that chose cracker A and the number of all surveyed people under 20. Let's find this ratio! 152/250 = 0.608 = 60.8 %

Subchapter links

9.2.1 Quadratic Applications with Inequalities p.502-503

9.2.2 Solving Systems of Equations p.507-508

Exercise 72 Page 508

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