Exercise 142 Page 315

a Probability is calculated by dividing the number of favorable outcomes by the number of possible outcomes.

P=Number of favorable outcomes/Number of possible outcomes From the exercise, we know that there is a total of 212 students in geometry. Of these, 64 are freshman and 112 are sophmores. Now we can determine the probabilities of selecting the different types of students. P(freshman)&=64/212 [0.8em] P(sophmore)&=112/212 To calculate the probability of selecting either a freshmen or a sophmore, we have to use the Addition Rule to add these probabilities and subtract the intersection of these sample spaces. P(A or B)=P(A) +P(B)-P(A and B) By substituting the values into the Addition Rule, we can calculate the probability of selecting either type of student. Notice that you cannot both be a sophmore and a freshman at the same time so there is no intersection between these sample spaces.

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

SubstituteValues

Substitute values

P(freshman or sophmore)=112/212 +64/212-0

SubTerm

Subtract term

P(freshman or sophmore)=112/212 +64/212

AddFrac

Add fractions

P(freshman or sophmore)=176/212

ReduceFrac

a/b=.a /4./.b /4.

P(freshman or sophmore)=44/53

The unusual part is that there was no overlap between the different sample spaces.

b To make things a little easier, we will introduce a couple of variables.

B&=performs in the band C&=performs in the chorus We want to know the intersection of students performing in a band and a chorus. To calculate this probability, we have to isolate P(B and C) in the Addition Rule. P(B or C)=P(B)+P(C)- P(B and C) From the exercise, we know that there is a 75 % chance a randomly selected geometry student is doing either band or chorus. Since there is a total of 212 students in geometry class, we have 0.75(212)=159 students from geometry in either band or chorus. With this, we can determine a few probabilites that we will need. P(B)&=114/212 [0.8em] P(C)&=56/212 [0.8em] P(B or C)&=159/212 Now we have enough information to calculate the intersection of B and C.

P(B or C)=P(B)+P(C)-P(B and C)

SubstituteValues

Substitute values

159/212=114/212 +56/212-P(B and C)

▼

Solve for P(B and C)

AddFrac

Add fractions

159/212=170/212-P(B and C)

SubEqn

LHS-170/212=RHS-170/212

159/212-170/212=- P(B and C)