Hello, I am studying for an exam on Monday. I am stuck on these 4 practice questions.
1) 60 students were asked about movies they had seen during the past month. 20 of the students had seen a comedy, 15 had seen a horror movie and 5 had seen both a comedy and a horror movie. Find the probability that a randomly selected student had seen a horror movie or a comedy or both.
P( CUH) = P(C) + P(H) - P (C∩H)
= 20+ 15 -15-5 =30?
2) A survey of 450 freshmen at a local community college showed that 100 were enrolled in a math course, 122 were enrolled in an english course, and 75 were enrolled in a history course. Furthermore, 52 freshmen were enrolled in both math and english, 42 were enrolled in both english and history, and 45 were enrolled in both math and history. Finally, 32 freshmen were enrolled in all three types of courses.
Do I just add what pertains to each circle? For example for History, 42+32+45
3) If P(EUF)=.9, P(E)=.6, P(E∩ F)=.1
a) Find P(F)= .3
I did P(EUF) = P(E) +P(F)-P(E∩ F)
=.9=.6+?=.1
=.3
B) Find P(Ecomplement)
= .2?
4) Given: n(u)=100, n(A∩ B) =30, n(AUB)=80, n(b)=50
a) n(Acomplement∩ Bcomplement) =
n(u) - n (A∩ B) =
100-30=70
b) n(B∩ ∅) What does this even mean?
Venn diagrams stump me a bit but I know that they are more useful and quicker than the addition rule.
For question 2, I am only supposed to set up the venn diagram.
imgur.com/a/u 8bdn9g
For reference to question 2 because this website doesn't allow urls.
without the space between u and 8
#1 number( CUH) = number(C) + numbr(H) - number (C∩H)
= 20 + 15 - 5
= 30
You had 20+ 15 -15-5 =30 , probably a typo in the -15
It asked for the prob P(CUH)
= 30/60 = 1/2
#2, Venn diagram are quite easy, especially when the data is nicely presented.
Draw 3 intersecting circles and label them M, E, and H
- start with the intersection of all 3, and place 32 in that intersection
- look at the M-E football-looking intersection, it is to contain 52, but we have already 32 of those accounted for. So in the remaining part of M-E put in 20.
- similarly, put 10 in the open part of H-E, and 13 in the open part of M-H.
Now look at the whole M circle, it said 100 took math, but the M circle already contains 13+32+20 or 65 , so 35 go into the Math-Only part
Do the same for the E and H circles, namely 60 in the E-only, and 20 in the H-only.
Now add them all up and compare it with 450 students.
If that sum is less than 450, that difference would be the students not taking any of the 3 subjects, if it is more than 450, then the question is bogus.
#3
You should present it like this:
P(EUF) = P(E) +P(F)-P(E∩ F)
.9 = .6 + P(F) - .1
.9 - .6 + .1 = P(F)
P(F) = .4
P( E complement) to me means ( not E)
so P( E complement) = 1 - .6 = .4
#4
from n(A∩ B) =30, n(AUB)=80, n(B)=50
n(AUB) = n(A) + n(B) - n(A∩ B)
80 = n(A) + 50 - 30
n(A) = 60, so n(notA) = 100-60 = 40 and n(notB) = 100-50 = 50
use your definitions to carry on
as to n(B∩ ∅) , the symbol ∅ is the 'NULL" set and
∅∩ anything is the null set
while ∅U anything is that anything.
1) To find the probability that a randomly selected student had seen a horror movie or a comedy or both, you can use the principle of inclusion-exclusion.
First, find the probability of watching a comedy movie (P(C)) and the probability of watching a horror movie (P(H)). In this case, P(C) = 20/60 = 1/3 and P(H) = 15/60 = 1/4.
Next, find the probability of watching both a comedy and a horror movie (P(C∩H)). In this case, P(C∩H) = 5/60 = 1/12.
Finally, use the formula P(C U H) = P(C) + P(H) - P(C∩H) to calculate the probability of watching either a comedy or a horror movie or both. In this case, P(C U H) = 1/3 + 1/4 - 1/12 = 11/12.
Therefore, the probability that a randomly selected student had seen a horror movie or a comedy or both is 11/12.
2) To find the number of freshmen enrolled in a specific course or combination of courses, you can use the principle of inclusion-exclusion.
For example, to find the number of freshmen enrolled in the history course, you need to add the number of freshmen enrolled in only history (75), the number of freshmen enrolled in both history and math (45), the number of freshmen enrolled in both history and English (42), and the number of freshmen enrolled in all three courses (32). So, the total number of freshmen enrolled in the history course is 75 + 45 + 42 + 32 = 194.
Similarly, you can find the number of freshmen enrolled in other courses or combinations of courses by adding the appropriate numbers.
3) a) To find P(F), we can use the formula P(EUF) = P(E) + P(F) - P(E∩F). We are given that P(EUF) = 0.9, P(E) = 0.6, and P(E∩F) = 0.1.
Substituting these values into the formula, we have 0.9 = 0.6 + P(F) - 0.1. Solving for P(F), we get P(F) = 0.9 - 0.6 + 0.1 = 0.4.
Therefore, P(F) = 0.4.
b) To find P(Ecomplement), we use the fact that the complement of an event E is the probability of not E happening, which is equal to 1 - P(E). Given that P(E) = 0.6, we have P(Ecomplement) = 1 - 0.6 = 0.4.
Therefore, P(Ecomplement) = 0.4.
4) a) To find n(Acomplement∩Bcomplement), we know that the complement of event A is all the elements in the universal set U that are not in A. So n(Acomplement) = n(U) - n(A). Similarly, n(Bcomplement) = n(U) - n(B).
Given that n(U) = 100, n(A∩B) = 30, and n(AUB) = 80, we can find n(A) and n(B) by using the principle of inclusion-exclusion.
n(AUB) = n(A) + n(B) - n(A∩B)
80 = n(A) + n(B) - 30
n(A) + n(B) = 110
Also, n(U) = n(AUB) + n(Acomplement∩Bcomplement)
100 = 80 + n(Acomplement∩Bcomplement)
n(Acomplement∩Bcomplement) = 100 - 80 = 20
Therefore, n(Acomplement∩Bcomplement) = 20.
b) n(B∩∅) represents the intersection of event B and the empty set (∅). The empty set (∅) contains no elements, so the intersection with any set will also be empty. Therefore, n(B∩∅) = 0.