<p>If I had a 30 percent chance of being accepted to five different colleges, what would be the chance of getting into one of the five? Into two? Three and so on?</p>
<p>How do you do this kind of problem?</p>
<p>If 30% is the chance of getting accepted in all of the five colleges, then the chance of getting is only one college is the fifth root of 30%. Suppose x = 5th-rad(.3), which is the chance of getting in one college. So the chance of getting in 2 out of 5 is x^2. 3 out of 5 is x^3, and so forth.
But those are just the average percentages, though, because usually we would multiply the individual chances of getting accepted at each college, which are likely different, and end up with the general chance of getting into all 5. Now we’re decomposing the general chance, so it just gives the average/predictable/expectational individual chances.</p>
<p>I might be wrong. Need some math-geek confirmation.</p>
<p>[thread=876372]UT Austin Cockrell Engineering Chances[/thread]
Chance me if you would, thanks.</p>
<p>I’ve determined that you mean “If the probability of being accepted by each of 5 colleges is .3, what is the probability you are accepted by 1 and only one college or 2 and only 2 colleges or 3 and only 3 and so on”. </p>
<p>To determine the probabilities it needs to be assumed that each of the college acceptance decisions are independent.</p>
<p>If you look at the case where you are accepted into 1 college and not accepted by 4 colleges, there are actually 5 different ways for this to happen (with each of the 5 colleges accepting you). So you would have .3<em>.7^4 which is the probability of being accepted by an individual college multiplied by the probability of not being accepted by each of the other 4 colleges. Since this can happen 5 different ways, you multiply this by 5: 5</em>.3*.7^4=.36015.</p>
<p>The cases where you are accepted into 2 colleges or 3 colleges are more difficult and require a discussion
If you are familiar with Pascal’s triangle you can find the binomial coefficients or you can use factorials and the formula nCk= n!/(k!(n-k)!).
[Binomial</a> distribution - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Binomial_distribution]Binomial”>Binomial distribution - Wikipedia)
[Pascal’s</a> triangle - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Pascal’s_triangle]Pascal’s”>Pascal's triangle - Wikipedia)</p>
<p>I’ve written enough here, and so will just say:</p>
<p>10<em>.3^2</em>.7^3=.3087 is the probability that you are accepted by 2 colleges
10<em>.3^3</em>7^2=.1323 is the probability that you are accepted by 3 colleges
5<em>.3^4</em>.7^1=.02835 probability for 4 colleges
.3^5=.00243 probability for 5 colleges</p>
<p>and you didn’t ask but the probability that you aren’t accepted by any college is
.7^5=.16807.</p>
<p>How did you do that? shouldnt the percents add to 100.</p>
<p>It’s interesting that you should inquire about this. I did something similar to calculate my admission probabilities at the six universities that I applied to after finding a realistic estimate of my chances (based on my academic statistics) for each individual college (which made what I did more complex and time-demanding, of course).</p>
<p>I’ll reiterate much of what bjohnson04 stated if it provides any additional clarification. Also, I will refer to these five universities as A, B, C, D, and E for purposes offering possible admission scenarios and to illustrate the concept without formulas that are not readily intuitive.</p>
<p>
</p>
<p>Were you confused by the coefficients that he used? They are used to account for the number of college decision possibilities in this hypothetical scenario, of which there are 32 in total (waitlists are assumed to be categorically equal to rejections).</p>
<p>For admission to zero (:(), there is only one way in which that may be accomplished – A, B, C, D, or E. Thus, the coefficient placed before the computation is one:</p>
<p>1 * 0.3 ^ 0 * 0.7 ^ 5 = .16807 = 16.807%</p>
<p>For admission to one university, there are five possibilities – University A, B, C, D, or E. Thus, we must multiply the probability result by 5 in order to obtain the probability:</p>
<p>5 * 0.3 ^ 1 * 0.7 ^ 4 = .36015 = 36.015%</p>
<p>For admission to two universities, there are ten possibilities – AB, AC, AD, AE, BC, BD, BE, CD, CE, or DE. Thus, we multiply the probability by ten:</p>
<p>10 * 0.3 ^ 2 * 0.7 ^ 3 = .3087 = 30.87%</p>
<p>For admission to three universities, there are once again ten possibilities – ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, or CDE. Thus, we multiply by ten:</p>
<p>10 * 0.3 ^ 3 * 0.7 ^ 2 = .1323 = 13.23%</p>
<p>For admission to four universities, there are five possibilities – ABCD, ABCE, ABDE, ACDE, or BCDE. Thus, we multiply by five:</p>
<p>5 * 0.3 ^ 4 * 0.7 ^ 1 = .02835 = 2.835%</p>
<p>To receive admission to all five, there is just one possibility, similar to the scenario of not receiving admission to any - A, B, C, D, or E. Thus, we multiply by one:</p>
<p>1 * 0.3^5 = .00243 –> 0.243%</p>
<p>
</p>
<p>They do accumulate to 100, as they should, since the entire range of circumstances is considered.</p>
<p>If you wish to think of this in terms of cumulative chances we can simply add the percentiles:</p>
<p>[ul] [<em>] Chances of admission to 5: 0.243%
[</em>] Chances of admission to 4 or more: 3.078%
[<em>] Chances of admission to 3 or more: 16.308%
[</em>] Chances of admission to 2 or more: 47.178%
[<em>] Chances of admission to 1 or more: 83.193%
[</em>] Chances of admission to 0 or more: 100% (Obviously) [/ul]</p>
<p>Ah I see.Thank you all.</p>