This question is very simple I think, but I thought somebody here could explain it to me better.
Here is the situation:
I am in a carpool with 2 other people.
In our jobs, we have to work 2 Sundays out of every 8 weeks. In this current 8 week span, one Sunday was Easter, so no one worked that day. So there were only 7 Sundays in question.
However, none of the 3 of us work the same Sunday. Assuming that we can be assigned randomly to any Sunday (we can’t, but the rules were not applied equally to each one of us), what are the odds that no two of us have any Sundays in common?
To sum up:
7 Sundays
3 people
each of the 3 has to work 2 Sundays
Say the people are Alice, Bob, and Cartwright. The number of possibile assignments for Alice alone are 7C2 = 21. Same with Bob and Cartwright. So, if we randomly assigned you each two Sundays, that’s 21[sup]3[/sup] = 9261 possibile setups.
But we have to discount some of these. We have to discount all the setups in which all three of you are working on the same Sunday. For instance, let’s count all the setups in which all three of you are working on Sunday #1. Alice is working also one of the other six Sundays, as are Bob and Cartwright. So that’s 6[sup]3[/sup] = 216. The same can be said for Sundays #2-#7, so that’s 7 × 216 = 1512. But, I counted some setups twice here. Say, the setup in which all of you work Sunday #1, and all of you work Sunday #2. There are 7C2 = 21 of these setups, so the total number of setups we need to discount is 1512 - 21 = 1491.
Thus the total number of possible setups is 9261 - 1491 = 7770.
Now let’s consider for how many of these no two of you is working the same day. In such a setup, one of the seven Sundays will have nobody working. Let’s assume it’s Sunday #1. That leaves 6 Sundays, out of which Alice must have two, and Bob must have two. Cartwright will have whatever is left over. There are 6C2 = 15 possiblities for Alice’s two dates. No matter which two they are, it leaves four available, so there are 4C2 = 6 possibilites for Bob’s two dates. Thus there are a total of 15 × 6 = 90 setups which nobody works together, and nobody works on Sunday #1. You can do the same for Sundays #2-#7, so there are a total of 7 × 90 = 630 setups for which nobody works together.
So, 7770 possible setups, all equally likely. 630 of these fulfill the condition. Thus the probability is 630/7770 = 3/37 = 0.08108.
I think I misread the problem. I thought you made the stipulation that the three of you could not all work on the same day. But I guess you didn’t. So, forgetting about that, the analysis is somewhat easier, and you can ignore my second and third paragraphs in the previous post. The answer is then 630 / 9261 = 10 / 147 = 0.06803.
Achernar, we can arrive at your answer without resorting to combinatorics:
A is allotted two days.
When B’s first day is alloted the chances of this falling on a non-A day is 5/7 (there are 5 “open” days out of seven).
When B’s second day is allotted the chances are 4/6 (there are 4 open days out of 6 (B’s first day fills one of the seven).
When B’s first day is alloted the chances of this falling on both a non-A day AND a non-B day is 3/7.
When C’s second day is allotted the chances are 2/6
We can muliply these independant probabilities:
5/7 x 4/6 x 3/7 x 2/6 = 0.68 (2dp)
I notice the actual allotment put each day in different 4-week halves – if this is a rule the probabilities becomes.
Allot A 2 days.
Allot B one non-A day in month 1, 3/4
Allot B one non-A day in month 2, 3/4
Allot B one non-A and non-B day in month 2, 2/4
Allot B one non-A and non-B day in month 2, 2/4
3/4 x 3/4 x 2/4 x 2/4 = 9/64 = 0.14 (2dp)
So BobT, I take it this really happened, you were alloted days, supposedly randomly, but it just turned out they were all different – and this is a drag because you none of you get to car pool? What a gyp! That sucks.
And replace some of those B’s with C’s, snaffen-raffen-dirty-cut-and-paste.
Dagnabbit – here is what I meant:
A is allotted two days.
When B’s first day is alloted the chances of this falling on a non-A day are 5/7 (there are 5 “open” days out of seven).
When B’s second day is allotted the chances are 4/6 (there are 4 open days out of 6 (B’s first day fills one of the seven)).
When C’s first day is alloted the chances of this falling on both a non-A day AND a non-B day are 3/7.
When C’s second day is allotted the chances are 2/6
We can muliply these independant probabilities:
5/7 x 4/6 x 3/7 x 2/6 = 0.68 (2dp)
I notice the actual allotment put each day in different 4-week halves – if this is a rule the probabilities becomes.
Allot A 2 days.
Allot B one non-A day in month 1, 3/4
Allot B one non-A day in month 2, 3/4
Allot C one non-A and non-B day in month 2, 2/4
Allot C one non-A and non-B day in month 2, 2/4
3/4 x 3/4 x 2/4 x 2/4 = 9/64 = 0.14 (2dp)
Admin – please take one off my post count, thank you.
I’m not even going to talk about the 2 that should be a 1, nor the idiosyncratic spelling of independent. In fact there are so many things I’m not going to
Though I admit I probably would have done it your way if I had been solving the right problem from the start.