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Probability. Why can't I add them?

  • Thread starter Dafydd
  • Start date
  • #1
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Homework Statement



Let's say, at a certain university, over the last few years 100 students have graduated with a degree in Computer Science, and another 100 in Biology. Let's also say 60 of the CS graduates found a job within 3 months of their graduation, and 50 of the Bio graduates did.

The question is: what is the probability that a student from either (as in "at least one of)) programme gets a job within 3 months of graduation?

Homework Equations



-

The Attempt at a Solution



First, I tried adding the groups together, saying the probability is 0.5 for Bio and 0.6 for CS, and slapped them together thinking the two are independent events, but this gave me a probabality of 1.1, which can't be right. So then I thought maybe the probability should be (50 + 60) / (100 + 100), which is 0.55, and this made sense to me because as a matter of fact, 110 out of 200 students did find a job within 3 months of graduation.

But then someone suggested the answer should be 1 minus the probability that none of them get a job, which would be 1 - ((1- 0.60) * (1- 0.50)) = 1 - (0.40 * 0.50) = 0.80 != 0.55. And then I thought maybe my first approach had overlooked the possibility of graduates from both programmes finding a job. But adding the following together:

The probability of a CS grad but not the Bio one finding a job 0.60 * 0.50
The probability of a Bio grad but not the CS one finding a job 0.40 * 0.50
gives 0.30 + 0.20 = 0.50 != 0.55.

I'm really confused here. What am I doing wrong? I seem to have misunderstood something very fundamental.
 

Answers and Replies

  • #2
1,953
248
If noone got a degree in both computer science and biology, the (50+60)/(100+100) answer is correct. if some of them did graduate in both, the problem is unanwerable.

We want to find the probablity that one randomly drawn student from the 200 graduates is did find a job. This is simply:
(total# of students that found a job)/(total # of students)


why everything else is wrong

- at first you talk about independent events, but you aren't very clear what those events are. Even if two events are independent, that won't make it right to add their
probabilities. If two events A and B are independent then P(Both A and B occur)
is P(A)P(B). If two events are exclusive then P(A or B occurs) = P(A) + P(B)

- We're talking about the probalities that a single student finds a job.
The possibilities are:
- it's a Bio student and did find a job
- it's a Bio student and did not find a job
- it's a CS student and did find a job
- it's a CS student and did not find a job

If you add the 2 probabilities for the case where the student did find a job you also
get 0.55
 
  • #3
12
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I forgot to thank you for your reply. I found it very helpful. So thank you.
 

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