SOLUTION: Suppose there is a group of 100 students. 70 of them are in an English class, and 40 of them are in a Math class (so there are 10 students in both). Suppose we select a singl

Question 1206041: Suppose there is a group of 100 students. 70 of them are in an English class, and 40 of them are in a Math class (so there are 10 students in both).
Suppose we select a single student at random. Let E= chosen student is in an English class and M=chosen student is in a Math class.
Compute the following probabilities.
P(M)=
P(E)=
P(M and E)=
P(M or E)=
P(M|E)=
Are these two events independent?

Now suppose we select two students at random. You can treat this like selecting one, then selecting another, without replacement.

What is the probability of selecting a student only in an English class first, and then selecting a student only in a Math class second (neither is in both)?

What is the probability of selecting a student only in a Math class second if my first selection is a student who is only in an English class?

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

You posted too many questions. Ideally you should post one problem at a time.
Multi-part questions are fine as long as they don't get too lengthy.

I'll answer the first block where it ends with "P(M|E)"

E = set of people taking English
M = set of people taking Math
n(E) = number of people taking English
n(E) = 70
n(M) = 40
n(M and E) = 10
n(M or E) = n(E)+n(M) - n(M and E) ... inclusion-exclusion principle
n(M or E) = 70+40-10
n(M or E) = 100

P(M) = probability a student takes math
P(M) = n(M)/n(total)
P(M) = 40/100
P(M) = 2/5

P(E) = n(E)/n(total)
P(E) = 70/100
P(E) = 7/10

P(M and E) = n(M and E)/n(total)
P(M and E) = 10/100
P(M and E) = 1/10

P(M or E) = n(M or E)/n(total)
P(M or E) = 100/100
P(M or E) = 1
A probability of 1 indicates 100% certainty that this event will happen.
This is because all 100 students either take English, Math, or both.

P(M | E) = P(M given E)
P(M | E) = P(M and E)/P(E)
P(M | E) = P(M and E) ÷ P(E)
P(M | E) = (1/10) ÷ (7/10)
P(M | E) = (1/10) * (10/7)
P(M | E) = 10/70
P(M | E) = 1/7
Another approach:
P(M | E) = n(M and E)/n(E)
P(M | E) = 10/70
P(M | E) = 1/7
P(M given E) means we know 100% that the student takes English.
We focus on the 70 English students.
Of these 70 students, 10 take math as well. So 10/70 = 1/7 is the conditional probability we're after.

Summary:
P(M) = 2/5
P(E) = 7/10
P(M and E) = 1/10
P(M or E) = 1
P(M|E) = 1/7

I'll give a hint about the next question.
The following equations are true if and only if both events are independent.

P(M given E) = P(M)
P(E given M) = P(E)
P(M and E) = P(M)*P(E)

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