Probability question

Question

A college offers two courses in Calculus - Calc 1 and calc 2. Last semester, 60.9% of students took calc 1 and the rest took calc 2 ( no student took both the courses). Of the students who took calc 1, 28% received A's and of the students who took calc 2, 38.1% received A's. If a student is selected from random from those who took A's, what is the probability that the student took calc 1?

 

a) 0.170

b)0.280

c)0.534

d)0.609

e)0.720

Answer 1

Hello Arch! The easiest method to obtain this answer is by constructing a 2 way table from the information provided so that you can utilize the conditional probability rules. Since there is no intersection between Calc 1 and Calc 2, these two subjects can be placed as your vertical column headings with the totals .609 and .392 placed at the bottom. The horizontal row headings would be earned an A or didn't earn an A.

Next, you'd multiply the percent of As for each subject by their total. Calc 1 (.69 times .28 = .17052) and Calc 2 (.381 times .392 =.149352). These answers would be placed in the row for As under their appropriate subject(vertical column). To get the amount of students that didn't have As, subtract the As in the subject from the total taking that subject. Calc 1 (.609 - 17052 = .43848) and Calc 2 (.392 - .149352 = .242648). So, your two way table would read Calc 1 and A (.17052), Calc 1 and no A (.43848), Calc 2 and A (.149352) and Calc 2 and no A (.242648). You can total the 2 way table up both ways to see that your grand total (bottom right) equals 1.0 or 100%.

After getting the two way table, the conditional probability is easy. The condition is the denominator and the other portion of the question is the numerator. Given that the student made an A (.391872), what is the probability that they took Calc 1 (.17052). So, place (.17052/.319872) which is .53308. Answer choice C is the correct answer, as the small differential can be explained due to the fact that I didn't round.

Have a good one now,

Dr. P

Answer 2

pick a number strategy just like on the SAT or ACT.

Suppose there are 100 students.

Then 60 in calc1 and 40 in calc 2.

28% of 60 is 16.8

38% of 40 is 15.2

Total is 32 grades of A

In light of this, probAbility looks like 1/2.

To make this legit and generalize, you suppose there are N students and repeat the same argument.