We throw 250 darts (one by one) at 1200 dart boards. Each dart strikes a board randomly. No dart's strike affects any other. What is the probability that two darts will strike the same board?
I make it 1 in 1200, b/c the probability of the 1st dart striking a board is 1. The probability of the 2nd dart striking the SAME board is 1/1200. Correct?
2nd part: When all 250 darts have been thrown, what is the probability that ANY two have struck the same board? Using the same reasoning: the first dart strikes one board out of 1200, the second 1 board out of 1 occupied and 1199 unoccupied (1:1199), 3rd 1 board out of 2 occupied/1198 unoccupied (2:1198), etc. This is not a factorial b/c we don't care which boards are hit -- they aren't unique. When 249 have been thrown, the odds that #250 will hit an occupied board are 249:951, or less than 25%. Correct?
For the first part of the question, you are correct. The probability that the second dart will strike the same board as the first dart is indeed 1/1200. This is because there are 1200 possible boards for each dart to strike, and they are all equally likely. So, the probability of both darts striking the same board is 1/1200.
For the second part of the question, you are on the right track, but your reasoning is not entirely correct. Let's break it down step by step.
- The first dart can strike any of the 1200 boards, so the probability is 1/1200 that it will hit a particular board.
- For the second dart, there are now two possibilities. It can either hit the same board as the first dart (probability 1/1200) or hit a different board (probability 1199/1200). So the overall probability that the first two darts hit the same board is (1/1200) + (1199/1200) * (1/1200).
- For the third dart, there are three possibilities. It can hit the same board as the first dart (probability 1/1200), the same board as the second dart (probability 1/1200), or a different board (probability 1198/1200). So the overall probability that any two of the first three darts hit the same board is [(1/1200) + (1199/1200) * (1/1200)] + (1198/1200) * (1/1200).
You can continue this reasoning for all 250 darts, adding up the probabilities for each possibility. The final result will be the probability that any two of the 250 darts hit the same board. This calculation can be computationally complex, but you can use the principle of inclusion-exclusion to simplify it. The exact probability depends on the number of boards and darts, but it will likely be a relatively small number.
Your reasoning for the first part is correct. The probability that the first dart strikes a board is 1 (since there are 1200 boards to choose from). Given that the first dart has already struck a board, the probability that the second dart strikes the same board is 1/1200, since there are still 1200 boards to choose from and only 1 of them is the same as the first dart.
Now, let's move on to the second part. The probability that the third dart strikes a board that is already occupied by one of the previous two darts is 2/1199. This is because there are now only 1199 boards to choose from (since one is already occupied) and 2 of them are the same as the first or second dart.
Continuing this pattern, the probability that the fourth dart strikes a board that is already occupied by one of the previous three darts is 3/1198. And so on.
So, the probability that any two darts out of the 250 will strike the same board can be calculated as:
(1/1200) + (1/1200) * (2/1199) + (1/1200) * (2/1199) * (3/1198) + ...
This is a product of probabilities where the fraction multipliers increase by 1 each time.
Calculating this sum for all 250 darts would be complex and time-consuming. However, we can approximate it using a well-known formula called the Stirling approximation.
According to the Stirling approximation, n! (n factorial) can be approximated by √(2πn)(n/e)^n, where e is the mathematical constant approximately equal to 2.71828.
Using this approximation, the probability that any two darts out of the 250 will strike the same board can be approximated as follows:
P ≈ √(2π(250))/(2.71828^250)
Calculating this approximation gives us a probability value that is extremely close to zero. Therefore, the probability of any two darts striking the same board is extremely low given the large number of darts and dart boards involved.