Mark:
In general, the probability of selecting exactly K of R items chosen from a set of N items is:
C(R,K) * C(N-R,R-K) / C(N,R)
Note that when K=R, the numerator is 1, and we have the probability for the original problem.
In this case, the answer is:
C(6,3) * C(30,3) / C(36,6) = 0.0417
The number of ways that 3 numbers can be chosen from 6 is 20. The number of ways that 3 numbers can be chosen from the remaining 30 is 4,060. Multiplying these 2 numbers we obtain 81,200, which is the total number of ways 6 numbers can be chosen from 36 so that exactly 3 numbers will match the winning six. There are 1,947,792 possible combinations of choosing 6 numbers from 36 so the probability of having a ticket that matches 3 of the winning numbers is 81,200/1,947,792 which equals .04168822954. This means that the odds are about 1 in 24.
Stormy:
The clock at the station at the end of the first leg was five minutes fast.
Good effort Stormy, this is how I see it, which due to the confusion amongst the boys has made me doubt the answer and you may be right.
The station clock is 3 minutes fast. The morning journey took 65 minutes, and the evening journey therefore took 52 minutes, and the train arrived 57 minutes after it should have left, that is, 3 minutes early.
Summing the powers of 2, using each power only once, any positive integer may be obtained. For example using the numbers 1, 2, 4, 8, 16, and 32 any number from 1 through 63 can be formed. For example, 9= 8+1, 10= 8+2, 11= 8+3, and so on.
(Remember that each power of 2 can be used only once in each sum and a sum such as 4 + 4 + 2 =10 would not be allowed).
So, given these rules, how would the number 1,000,000 be obtained