3rd try: i applied for 3 summer jobs. NON-RESIDENTS ONLY QUALIFY FOR ONE POSITION PER JOB bit not guaranteed 1

please give me the odds (this means like 42 to 1) of getting a position on ANY job- qualifying for one job does not make the recipiant ineligible for the other jobs. June job has 6 positions : 80 non-res applicants %26amp; 20 non-res (i have a 100 to 1 chance. then if a resident got the 1st position i have a 99 to 1 chance for position 2 ...then 98..etc. if a NON-resident got the 1st position i am inelleigible for the remaining 5 positions in June). July job has 5 positions: 70 residents apply %26amp; 30 non-res. once a non-res gets a July position the other non-red are ineligible for a July position. August job has15 positions: 1,200 resident applications %26amp; 800 non-residents. once a non-res gets a position the other non-res are removed from the eligible applicants. NON-RES ARE NOT GUARANTEED A POSITION, they are only eligible for 1 position per job 1) please give me the non-res odds per job. 2) please give me the odds of a non-res getting ANY pos 3) please give me the formulas.

3rd try: i applied for 3 summer jobs. NON-RESIDENTS ONLY QUALIFY FOR ONE POSITION PER JOB bit not guaranteed 1
C(n, k) (pronounced "n choose k") is how many different k-element subsets an n-element set has, 0 %26lt;= k %26lt;= n. Numerically it is C(n, k) = n! / ((k! (n - k)!).





(If the order matters, the k items in the subset can be permuted in k! different ways, yielding the formula P(n, k) ["P" for "permutation"] = k! C(n, k) = n! / (n - k)!.)





Possible results for June:


RRRRRR: C(80, 6) ways


RRRRRN: C(80, 5) * C(20, 1) ways





for July:


RRRRR: C(70, 5) ways


RRRRN: C(70, 4) * C(30, 1) ways





for August:


15 R: C(1200, 15) ways


14 R, 1 N: C(1200, 14) * C(800, 1) ways





If there is a specific resident you are interested in, you can say, for example, for June, that instead of 80 R's there are 79 R's and one S (for specific). Or for one specific nonresident then instead of 20 N's Jine had 19 N's and one S.





So redoing for a specific nonresident S:





June:


RRRRRR: C(80, 6) ways = 300,500,200


RRRRRN: C(80, 5) * C(19, 1) ways = 456,760,304


RRRRRS: C(80, 5) * C(1, 1) ways = 24,040,016


If all ways are equally likely, the probability that the specific nonresident "S" got a June job is: 24,040,016 / (300,500,200 + 456,760,304 + 24,040,016) = 24,040,016 / 781,300,520 = 2/65. Where probability is like A/(A + B), the corresponding odds are A to B ... here that is 24,040,016 to (300,500,200 + 456,760,304) or the same thing 2 to 63 or 1 to 31.5 in favor of "S" getting a June job, or 31.5 to 1 against. (So probability = 2/65 = 1/32.5 or 1 *in* 32.5, making the odds 1 *to* 31.5 for.)





July:


RRRRR: C(70, 5) ways = 12103014


RRRRN: C(70, 4) * C(29, 1) ways = 26589955


RRRRS: C(70, 4) * C(1, 1) ways = 916,895


prob: 1/43.2, odds for 1:42.2





August:


15 R: C(1200, 15) ways = rather large :-)


14 R, 1 N: C(1200, 14) * C(799, 1) ways


14 R, 1 S: C(1200, 14) * C(1, 1) ways


The numbers work out to:


prob: 15 / 13186, odds for 15 to 13171 or 1 to 878+(1/15)





If odds for 1 to x, then odds against are x to 1.





If p, q, r are the probabilities of getting Jun - Jul - Aug, then the probability of not getting any of them is (1 - p)(1 - q)(1 - r) and so the probability of getting at least one of them is 1 - ((1 - p)(1 - q)(1 - r)) = 1 - (1 - p - q - r + pq + pr + qr - pqr) = p + q + r - pq - pr - qr + pqr (which is close to but not quite p + q + r if p, q, and r are all very small). Anyway, here that works out to a combined probability that specific nonresident "S" gets at least one of the jobs is 1,116,593 / 20,570,160 or approx. 5.43%. The corresponding odds are 1,116,593 to 19,453,567 or approx. 1 to 17.42 for (or 17.42 to 1 against).





Dan
Reply:non-residents??





residents??





DISCRIMINATION!!!