# Thread: Dependent Events

1. ## Dependent Events

Morgan believes she has a 60% chance that she will be late for class if she misses the 8:00 bus. There is a 80% chance that she will miss the 8:00 bus on any day.

A) What is the probability that she will miss her bus tomorrow, but still be on time?

B) If there is a 90% chance that Morgan will be late, even if she catches the 8:00 bus, what is the probability that she will be late on any given day.

I know the formula I need to use, I just don't know which percentages go in which part of the formula? Would for part A this be correct??

p(A) =
p(B) = 80%
p(AandB)= 60%

p(AandB) = p(A) x p(B)
p(A) = 0.60/0.80
p(A) = 0.48 = 48%

2. ## Re: Dependent Events

Originally Posted by momofmaxncoop
Morgan believes she has a 60% chance that she will be late for class if she misses the 8:00 bus. There is a 80% chance that she will miss the 8:00 bus on any day.
A) What is the probability that she will miss her bus tomorrow, but still be on time?
B) If there is a 90% chance that Morgan will be late, even if she catches the 8:00 bus, what is the probability that she will be late on any given day.
Let $\displaystyle L$ be the event of being late to class.
Let $\displaystyle M$ be the event of missing the bus.
The statement “a 60% chance that she will be late for class if she misses the 8:00 bus” seems to me to mean “given that the bus is missed, she will be late for class”. $\displaystyle \mathcal{P}(L|M)=0.6$
That is the way that I read the “If…then…” form.

So part a) asks $\displaystyle \mathcal{P}(L\cap M)$

3. ## Re: Dependent Events

Originally Posted by momofmaxncoop
Morgan believes she has a 60% chance that she will be late for class if she misses the 8:00 bus. There is a 80% chance that she will miss the 8:00 bus on any day.

A) What is the probability that she will miss her bus tomorrow, but still be on time?

B) If there is a 90% chance that Morgan will be late, even if she catches the 8:00 bus, what is the probability that she will be late on any given day.

I know the formula I need to use, I just don't know which percentages go in which part of the formula? Would for part A this be correct??

p(A) =
p(B) = 80%
p(AandB)= 60%
No, you were told that P(A)= 80% and P(A|B) (probability A happens given that B happens) is 60%. P(A and B) = P(A|B)*P(B). However, "A" here is "she will be late for class". You want P(A* and B) where A* is the complement of A, the probability that she is NOT late for class. If P(A|B)= 80% what is P(A*|B)?

p(AandB) = p(A) x p(B)
p(A) = 0.60/0.80
p(A) = 0.48 = 48%

4. ## Re: Dependent Events

Originally Posted by HallsofIvy
No, you were told that P(A)= 80% and P(A|B) (probability A happens given that B happens) is 60%. P(A and B) = P(A|B)*P(B). However, "A" here is "she will be late for class". You want P(A* and B) where A* is the complement of A, the probability that she is NOT late for class. If P(A|B)= 80% what is P(A*|B)?
Halls you seem to be agreeing with my reading of this. But why?
Mind you I think we are both right.
I just cannot find a definitive justification of it

5. ## Re: Dependent Events

@Plato... I too agree with your statement The statement “a 60% chance that she will be late for class if she misses the 8:00 bus” seems to me to mean “given that the bus is missed, she will be late for class”. But, won't the A part of the question ask for the p(A*|B) , that is the probability of the student being on time despite missing the bus, ?

6. ## Re: Dependent Events

Originally Posted by MAX09
The statement “a 60% chance that she will be late for class if she misses the 8:00 bus” seems to me to mean “given that the bus is missed, she will be late for class”. But, won't the A part of the question ask for the p(A*|B) , that is the probability of the student being on time despite missing the bus, ?
That is correct. Part a) wants the value of $\displaystyle P(L^*\cap M)$
But that is $\displaystyle P(L^*|M)P(M)$.

From part b) we are given $\displaystyle P(L|M^*)=0.9$.
Then asked to find $\displaystyle P(L)$.

Note that in general $\displaystyle P(H^*|K)=1-P(H|K)$.
BUT $\displaystyle P(H|K^*)\ne 1-P(H|K).$

7. ## Re: Dependent Events

Originally Posted by Plato
I am posting this because I simply do not know how one reads the English here.
Let $\displaystyle L$ be the event of being late to class.
Let $\displaystyle M$ be the event of missing the bus.
The statement “a 60% chance that she will be late for class if she misses the 8:00 bus” seems to me to mean “given that the bus is missed, she will be late for class”. $\displaystyle \mathcal{P}(L|M)=0.6$
That is the way that I read the “If…then…” form.

One the other hand, it seems reasonable to argue that the phrase means and, as in $\displaystyle \mathcal{P}(L\cap M)=0.6$

Does anyone have a point view on this difference?
If so, can you support that point of view?
I was reading it exactly like you were, that is why I was asking for help and feedback on how about going on to answer the question.

8. ## Re: Dependent Events

Originally Posted by momofmaxncoop
I was reading it exactly like you were, that is why I was asking for help and feedback on how about going on to answer the question.