Derek D. answered • 06/25/20
Derek: Algebra/Statistics Tutor
A: Find the probability that a student is male: To rewrite this in probability notation, we would say P(male). For probability problems, we're often looking for a condition and a sample space. In this case, our sample space would be the total number of students: 72. The number of students matching our condition (male) is 34, since there are 34 total male students. Notice how part A doesn't care about their grades. Now, we make a fraction with the condition on top (the numerator) and the sample space on bottom (the denominator): 34/72. We can reduce this fraction (17/36) or divide 34 by 72 and write as a decimal (.4722) or rewrite the decimal as a percentage (47.22%).
B: Find the probability the student was Male AND got a "C": P(Male AND C): Our condition here involves two things at the same time, so the top half of our fraction is 15, since that's the number of students that both are male and have C's. Based on the wording for this part, we're still randomly selecting out of all students, so our sample space is 72. So our answer is: 15/72, or 5/24, or .2083, or 20.83%.
C: Find the probability the student was Male OR got a "C": P(Male OR C): Here, our condition is kinda flexibile. We're wondering what the chances are of the student, selected at random, being either a male or getting a C (or both. We're not picky.) How many students match this condition? Well, there are 34 males (if we add up that row) and there are 24 C's (if we add up that column). We want to make sure that we don't double count the students who are both Male and C though, so we'll add those together but subtract the 15 students that are at the intersection of Male and C. So we have: 34 (males) + 24 (C's) - 15 ( both). 34 + 24 - 15 = 43. That's the top half of our fraction, 43. We then divide by our sample space (which is still 72). So our answer is 43/72 (which can't be reduced), or .5972, or 59.72%.
D: If one student is chosen at random, find the probability they got an 'A' GIVEN they are female: P(A | female): The keyword here is 'Given'. 'Given' means we have additional information about our selection past just picking one at random. This affects our sample space, which goes on the bottom half of our fraction. In this case, instead of selecting out of the total, we select out of the females, of which there are 38. One way to think of this is that if we assume we're picking a female, we can ignore everything else, which means we're only focusing on those 38 females. Out of the females, how many match our condition? There are 18 females that have 'A's. So that makes 18/38, or 9/19, or .4737, or 47.37%.
I hope that helped!
