There are two more assignments in your course before the semester ends, and if you get an A on at least one of them, you will get an A for the semester. Your "what-if" scenario looks like this:

Event Probability

A on paper and A on presentation 0.25
A on paper only 0.10
A on presentation only 0.30
Do not get an A on either 0.35

Answer the following questions: (1 point each, except the last item, 2 points)

a. What is the probability of getting an A on the paper?
b. What is the probability of getting an A on the presentation?
c. What is the probability of getting an A in the course?
d. Are the grades on the assignments independent? Explain your answer

Step-by-step explanation:

Hello!

Given the events:

A: "obtain an A in the paper assigment" ⇒ P(A)= 0.10

B: "obtain A in the presentation assigment" ⇒ P(B)= 0.30

Ac: "do not get an A in the paper assigment" ⇒ P(Ac)= 1 - 0.10= 0.9

Bc: "do not get an A in the presentation assigment" ⇒ P(Bc)= 1 - 0.30= 0.7

P(A∩B)= 0.25

P(Ac∩Bc)= 0.35

a. P(A)= 0.10

b. P(B)= 0.30

c. To get an A in the course you have to get an A in the paper and an A in the presentation, symbolically: P(A∩B)= 0.25

d. If the events A and B are independent, then the ocurrence of one of them doesn't affect the probability of the other ocurring, if this is so then:

P(A)= P(A/B)

P(A) ≠ P(A/B) this means that "B" ocurring modifies the probability of ocurrence of A, i.e. both events are not independent.

I hope it helps!

step-by-step explanation:

a cube has 6 sides

the probability of getting 5 on the first cube is 1/6 since only one side of the cube has 5 on it.

there are 3 sides of the cube with even number on it [2,4,6]

therefore the probability of getting even number is 3/6

for both events to happen together, the probability would be

1/6 x 3/6 = 12/36

which simplifies to 1/12