# Thread: Probability of being late (where did I go wrong?)

1. ## Probability of being late (where did I go wrong?)

A maths lecturer called Dr. X, is occasionally late for lectures. Dr. X has two units, 1030 and 2051 and he is late 10% of the time for 1030 and 15% of the time for 2051. You overhear two students remaking that the good Dr was late again. What is the probability that this remark came from an 1030 student? You can assume that 75% of Dr. X's students are in 1030, 25% are in 2051 and that no student takes both units.

Here is what I did but I got a probability of 3.19 which is definitely wrong. Can someone find out what I did wrong? As I can't find it, thanks

2. Originally Posted by usagi_killer
A maths lecturer called Dr. X, is occasionally late for lectures. Dr. X has two units, 1030 and 2051 and he is late 10% of the time for 1030 and 15% of the time for 2051. You overhear two students remaking that the good Dr was late again. What is the probability that this remark came from an 1030 student? You can assume that 75% of Dr. X's students are in 1030, 25% are in 2051 and that no student takes both units.

Here is what I did but I got a probability of 3.19 which is definitely wrong. Can someone find out what I did wrong? As I can't find it, thanks
I like how you mentioned explicitly that we can assume the student's remark was correct; I can just imagine a devious professor adding on probabilities that certain students tell the truth, or have correct knowledge of the truth.

I also like your rigorous and formal set theoretic approach, but I prefer in this case to be more informal.

As you mentioned, we can assume the events are independent, and that there was randomness to the event of overhearing students, e.g., the listener wasn't hanging out in a place exclusively frequented by 1030 students.

Perhaps it's due to my having a small brain, but it seemed easiest for me to reduce the problem to a completely different but equivalent problem. Say we have 80 geese. We break them into two groups, one with 60 geese and one with 20 geese. Of the 60 geese, we mark 10% of them, that is, 6 of them. Of the 20 geese, we mark 15% of them, that is, 3 of them. We then release them into the wild. Later we find a goose that's marked. What's the probability that the goose was originally in the group with 60 geese?

It should be immediately clear that the probability is 6/9 = 2/3.

Edit: Actually I'm not as confident now as a moment ago that the problem is equivalent. I'll leave it up for consideration but if it's in error I'll try to correct it.

3. hmmm alright thanks for that.

However I'm still unsure of where I went wrong and how I can fix it.

4. Originally Posted by usagi_killer
hmmm alright thanks for that.

However I'm still unsure of where I went wrong and how I can fix it.
Ah yes, I figured that if my answer were correct, it would be easy to see how to translate it into formal language, and then the discrepancy would become clear. Throwing doubt on the correctness of my answer didn't help matters...

But I think I've determined a valid formal presentation, and it verifies the answer obtained in my first post.

I believe it is more proper to say "event" or "set of outcomes" rather than "set of events."

Let Bob be a randomly chosen student (the one that gets overheard).

Let $\displaystyle X_1$ be the event that Dr. X is late for Bob's class.
Let $\displaystyle X_2$ be the event that Dr. X is on time for Bob's class.
Let $\displaystyle A_1$ be the event that Bob is from unit 1030.
Let $\displaystyle A_2$ be the event that Bob is from unit 2051.

We are looking for $\displaystyle P(A_1|X_1)$.

We are given that

$\displaystyle P(A_1)=0.75$
$\displaystyle P(A_2)=0.25$
$\displaystyle P(X_1|A_1)=0.1$
$\displaystyle P(X_1|A_2)=0.15$

I attached a decision tree made with MS paint, a bit of an eyesore, but you get the idea. We have

$\displaystyle P(A_1 \cap X_1) = 0.75 \cdot 0.1$
$\displaystyle P(A_1 \cap X_2) = 0.75 \cdot 0.9$
$\displaystyle P(A_2 \cap X_1) = 0.25 \cdot 0.15$
$\displaystyle P(A_2 \cap X_2) = 0.25 \cdot 0.85$

Now we write

$\displaystyle P(A_1|X_1)$

$\displaystyle = \frac{P(A_1 \cap X_1)}{P(X_1)}$

$\displaystyle = \frac{P(A_1 \cap X_1)}{P((X_1 \cap A_1) \cup (X_1 \cap A_2))}$

$\displaystyle = \frac{P(A_1 \cap X_1)}{P(X_1 \cap A_1) + P(X_1 \cap A_2)}$

$\displaystyle = \frac{0.75 \cdot 0.1}{0.75 \cdot 0.1 + 0.25 \cdot 0.15}$

$\displaystyle = \frac{2}{3}$

Note that this is the exact same reasoning for finding out whether someone has a disease given that they test positive for a disease, when we are given probabilities for false positive and false negative. See this example on Wikipedia.

5. You are so awesome! Thanks so much!!