Something to think about: Probabilities!

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Hednej

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So all my friends are out playing soccer right now and I'm just driving myself crazy but can't figure this out.
I still have to hear back from some schools, and I'm sure a lot of you folks out there are waiting to hear from some schools too.
Let's say I am waiting for three schools.
According to MSAR, the ratio of people who got accepted from the interview pool are for school
1. 30%
2. 25%
3. 20%

So whats the probability that I will get into ONE of those schools? Not all, but just one, any one.

Now if I knew a math forum I would post it there, but I'm sure there are some math wizards amongst us.
 

Stolenspatulas

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So all my friends are out playing soccer right now and I'm just driving myself crazy but can't figure this out.
I still have to hear back from some schools, and I'm sure a lot of you folks out there are waiting to hear from some schools too.
Let's say I am waiting for three schools.
According to MSAR, the ratio of people who got accepted from the interview pool are for school
1. 30%
2. 25%
3. 20%

So whats the probability that I will get into ONE of those schools? Not all, but just one, any one.

Now if I knew a math forum I would post it there, but I'm sure there are some math wizards amongst us.

its been a while but i believe you have to compute 3 different cases.
A) getting into 1 and not 2 and 3 = (.3)(.75)(.80)
B) getting into 2 and not 1 and 3 = (.7)(.25)(.80)
C) getting into 3 and not 1 and 2 = (.70)(.75)(.20)

Then, add them all up.
 

Hednej

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getting into at least 1 of those 3: 58%
Can you tell us how you got 58?

its been a while but i believe you have to compute 3 different cases.
A) getting into 1 and not 2 and 3 = (.3)(.75)(.80)
B) getting into 2 and not 1 and 3 = (.7)(.25)(.80)
C) getting into 3 and not 1 and 2 = (.70)(.75)(.20)

Then, add them all up.

I don't know why but this one just makes sense and it came out to 46 or smt like that so I think your method's pretty reasonable. Thanks!
 

iwannagomed

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getting into school 1: 30%
getting into school 2: 25%
getting into school 3: 20%

NOT getting into school 1: 70%
NOT getting into school 2: 75%
NOT getting into school 3: 80%

NOT getting into school 1 & NOT getting into school 2 & NOT getting into school 3: (.70)(.75)(.80) = 0.42 = 42%

getting no acceptances from the three schools: 42%

GETTING AT LEAST ONE ACCEPTANCE (either 1, 2, or all 3): 58%

u can trust me on this one. i teach this stuff professionally.

but u shouldnt count on these numbers in real life. depending on ur credentials, it could be 0% or 100%.

be positive and keep ur hopes up~
just go out and play some soccer!!! ^^
 

weathertalk

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The probability that you won't get into any of those schools = 1-(probability you get into none of them). If we assume that acceptance/rejection from each school is an independent process then P(you don't get into any of them) = P(rejection at school A) x P(rejection at school B) x P(rejection at school C)
P(rejection) at any given school is, of course 1-p(rejection post interview).
 

sejin8642

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I did the simple probability calculation for you. Here's the answer.

No.1 school O O O X X X X X X X (30%)

No.2 school O X X X (25%)

No.3 school O X X X X (20%)

O represents acceptance, X represents rejection. So if you wanna know your chances of getting accepted to at least one school, draw lines through one letter of each school that at least include one O, count them, and divide it by 200 and multiply 100%.

So far you have 116 lines. Therefore, your chances of getting accepted is 58%.
 

dynx

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ummm...this isn't an appropriate question for analysis. The probablility could be calculated for random unrelated events but since the interview process alters the selection process knowing these numbers doesnt help. Math wise weathertalk was right.
 

DoctorB

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So all my friends are out playing soccer right now and I'm just driving myself crazy but can't figure this out.
I still have to hear back from some schools, and I'm sure a lot of you folks out there are waiting to hear from some schools too.
Let's say I am waiting for three schools.
According to MSAR, the ratio of people who got accepted from the interview pool are for school
1. 30%
2. 25%
3. 20%

So whats the probability that I will get into ONE of those schools? Not all, but just one, any one.

Now if I knew a math forum I would post it there, but I'm sure there are some math wizards amongst us.

I believe the answer is

3!
--------- (.3) (.25) (.2) = I don't have a calculator on me
1! 1! 1!
 

DoctorB

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It's additive. 75%.


No it isn't. You need to use factorials because the question is asking the odds of getting into ONE school. It would be additive if it were asking the odds of getting into AT LEAST one school. Multiplication if it were the odds of getting into all three. Factorials are used when you want to know the odds of ONE event happening.
 
D

deleted128318

It's additive. 75%.

If it were additive and you added another school to the list, with say, a 26% chance of being accepted, would your chance of receiving an acceptance be 101%?
 
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sejin8642

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If it were additive and you added another school to the list, with say, a 26% chance of being accepted, would your chance of receiving an acceptance be 101%?

I like the words "101% acceptance." :)
 

Doctor~Detroit

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So all my friends are out playing soccer right now and I'm just driving myself crazy but can't figure this out.
I still have to hear back from some schools, and I'm sure a lot of you folks out there are waiting to hear from some schools too.
Let's say I am waiting for three schools.
According to MSAR, the ratio of people who got accepted from the interview pool are for school
1. 30%
2. 25%
3. 20%

So whats the probability that I will get into ONE of those schools? Not all, but just one, any one.

Now if I knew a math forum I would post it there, but I'm sure there are some math wizards amongst us.

whoa, let's take a step back. you used msar for these percentages? msar only lists the number matriculated, *not* the number accepted. so if you calculated #matriculated/#applied for the percentages, you're going to underestimate your chances at each school. the percentages should be higher.
 

Law2Doc

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ummm...this isn't an appropriate question for analysis. The probablility could be calculated for random unrelated events but since the interview process alters the selection process knowing these numbers doesnt help. Math wise weathertalk was right.

Agree with this. This isn't a lottery and chance isn't controlling. There are things in the process you may have done that can help or hurt your chances, and there are subjective factors in the minds of adcoms in terms of "good fit" at work (i.e. the decks may be stacked for or against you). So statistics doesn't help here one iota. If you did poorly at your interviews, or adcoms felt you weren't a "good fit", you could have a zero percentage chance of getting in anywhere, regardless of the MSAR figures. Conversely, if the adcoms thought you were awesome, and you performed well, you could get all three. And this same set of issues applies at each school differently.
 

Doctor~Detroit

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Agree with this. This isn't a lottery and chance isn't controlling. There are things in the process you may have done that can help or hurt your chances, and there are subjective factors in the minds of adcoms in terms of "good fit" at work (i.e. the decks may be stacked for or against you). So statistics doesn't help here one iota. If you did poorly at your interviews, or adcoms felt you weren't a "good fit", you could have a zero percentage chance of getting in anywhere, regardless of the MSAR figures. Conversely, if the adcoms thought you were awesome, and you performed well, you could get all three. And this same set of issues applies at each school differently.

yeah, but if we accept this then we'd have one less thing to talk about on sdn! fun outweighs truth.
 

Law2Doc

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yeah, but if we accept this then we'd have one less thing to talk about on sdn! fun outweighs truth.

Fair enough. In that case, since we are presuming the Bizarro world where a pure coin flip decides admissions (there are only two choices at each school -- get in or not), he would have a 50% chance at each school, regardless of what the MSAR says.:rolleyes:
 

Doctor~Detroit

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Fair enough. In that case, since we are presuming the Bizarro world where a pure coin flip decides admissions (there are only two choices at each school -- get in or not), he would have a 50% chance at each school, regardless of what the MSAR says.:rolleyes:

well, flippin coins ain't fun, either. but using data to figure out #accepted/#applied, then (speciously) applying that as one's own probability for a particular school, and then figuring out the chance of getting into at least one school, can pass some time.

for myself, i only did this post-interview (i used #accepted/#interviewed for the schools i interviewed at). but then, i think i interview pretty well. in the end, it helps reassure people waiting to hear back from schools. some go to church, others crunch numbers.
 

notdeadyet

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These calculations are only valid on the assumption that each interviewee interviews exactly the same.

If the applicant nails each interview, his odds go up dramatically. If he bombs, his odds go down dramatically.

Also, if he's denied at school #1, his odds for schools #2 and #3 should probably go down. Folks who get shot down at one school are more likely to get shot down at another. Folks accepted at one are more likely to get accepted at another.
 

ltrain

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Fair enough. In that case, since we are presuming the Bizarro world where a pure coin flip decides admissions (there are only two choices at each school -- get in or not), he would have a 50% chance at each school, regardless of what the MSAR says.:rolleyes:

They're weighted coins because each school doesn't have room for 50% of its interviewees ;)

Assuming independence (which, like others note, clearly doesn't exist here), the 58% from above is right.
 

Stolenspatulas

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I never took stat, but i still think im right.

Remember the OP said he wants to know the prob of getting into ONE SCHOOL (and NOT the others).
 
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