There are 18 people in a team. 7 share the same month of birth(not the exact date). What is the probability of 7 out of 18 having the same birth month?
+118608 12*nCr(18,7)*(1/12)^7*(11/12)^11 = 0.0040925417608967
That is the probability of getting 7 in the same month but it includes the probability of getting 7 in 2 different months as well. I mean it includes it twice.
So you would have to subtract
18C7 * (1/12)^7 * (11/12)^11 * 11C7 * (1/12)^7 * (11/12)^4 * 12C2
nCr(18,7)*(1/12)^7*(11/12)^11*nCr(11,7)*(1/12)^7*(11/12)^4*nCr(12,2)=0.0000001463682172
I'm reasonably confident on the first part but have less confidence on the secontd part (the part to be subrtracted)
0.0040925417608967-0.0000001463682172 = 0.0040923953926795
Hi Melody:
What do you think of this approach:
First person ==Any month ==12/12
2nd person different from the 1st ==11/12
3rd person different from the first two ==10/12
.
.
And so on to the 12th person ==1/12
So, the probability that the first 12 persons have a different month each is: 12! / 12^12 ==0.0000537232 1709
Or: The probability that at least 2 share the same month is: 1 - 0.0000537232 1709==0.9999462768 - almost a certainty!! Now, I'm stuck on the remaining 6 !!. If the first 12 have each a different month, then the 13th will have 100% chance of sharing his/her month with one of the first 12!! What about the 14th person, and the 15th, 16th, 17th and 18th??
(18 nCr 7 * 11^11) + (18 nCr 8*11^10) + (18 nCr 9*11^9) +(18 nCr 10*11^8) +(18 nCr 11*11^7) + (18 nCr 12*11^6) + (18 nCr 13 * 11^5) + (18 nCr 14 * 11^4) + (18 nCr 15 * 11^3) + (18 nCr 16 * 11^2) + (18 nCr 17 * 11^1) + (18 nCr 18 * 11^0)==
(907975 8605524464, 113496 9825690558, 11464 3416736420, 937 9915914798, 62 0159729904, 3 2887258404, 1379884968, 44801460, 1086096, 18513, 198, 1, 0, 0, 0, 0, 0, 0, 0)==1033940 6236645784 /12^18== 0.000388358 - probability of at least 7 out of 18 people.
(18 nCr 7) * (11^11)==9079758605524464 / 12^18== 0.000341045 - probability of exactly 7 out of 18 people.
