# Math Help - Stuck on a few problems

1. ## Stuck on a few problems

I am stuck on a few probability problems. Any help would be greatly appreciated. I put the answers I came up with, but I'm not sure if they are correct.

1) many universities require a student to take a math placement test as well as a computer science placement test. 80% of the students who take both tests pass either the math or the science test. 60% pass the math, 40% pass the science. What is the probability that the student will pass both?

A=Pass Math
B= Pass Science

P(A or B)= P(A)+P(B)-P(A and B)

P(A and B)= P(A)+P(B)-P(A or B)

P(A and B)= .6+.4-.8= .2

This indicates that nobody fails though. I guess i'm not entirely sure if he's saying that 60% of the 80% who pass on or the other pass math, or just that 60% in general pass math.

Other way would be

P(A and B)
P(A)P(B)
(.48)(.32)

2)Plant A produces 70% of the floppy disks produced by Computec and plant B produces the other 30%. 1% of those produced by plant A have a flaw, 2% produced by plant B have a flaw. What is the probability that a floppy disk produced by Computec has a flaw.

I said
A= Plant A flaw
B= Plant B flaw
P(A)=(.01)(.7)
P(B)=(.02)(.3)

P(A or B)= P(A)+P(B)= .013

3)Jason owns two stocks. There is an 80% probability that stock A will rise in price, while there is a 60% probability that stock B will rise in price. There is a 40% chance that both stocks will rise in price. Are the stocks independent?

I said
A= Stock A rises
B= Stock B rises
P(A)=.8
P(B)=.6
P(A and B)= .4

So,

P(A and B)= P(A)P(B/A)
rearranging
P(A and B)/P(A)=P(B/A)

.4/.8=.5

.5 does not equal .6 so they are not independent.

2. Hello, kittysnyde!

Thank you for showing your work!

1) A university requires a student to take a math placement test
and a computer science placement test.
80% of the students who take both tests pass either the math or the science test.
60% pass the math, and 40% pass the science.
What is the probability that the student will pass both?

$\text{We have: }\:\underbrace{P(M \cup S)}_{0.8} \;=\;\underbrace{P(M)}_{0.6} + \underbrace{P(S)}_{0.4} - P(M \cap S)$

. . . . . $0.8 \;=\;0.6 + 0.4 - P(M \cap S) \quad\Rightarrow\quad P(M \cap S) \;=\;0.6 + 0.4 - 0.8$

Therefore: . $P(M \cap S) \;=\;0.2 \;=\;20\%$

3) Jason owns two stocks.
There is an 80% probability that stock A will rise in price,
while there is a 60% probability that stock B will rise in price.
There is a 40% chance that both stocks will rise in price.
Are the stocks independent?
Let: $A$ = stock A rises, $B$ = stock B rises.

If the events are independent, then: . $P(A \cap B) \;=\;P(A)\cdot P(B)$

. . We are given: . $P(A \cap B) \:=\:{\color{blue}0.40}$

. . And: . $P(A)\cdot P(B) \:=\:(0.80)(0.60) \:=\:{\color{blue}0.48}$

No, they are not independent.

3. Thank you very much!