Let's start with two objects A and B. There are obviously 2! (=2) arrangements of them. In one arrangement, both objects are in their natural place, and in the other neither is. So if we use p(n) to denote the probability that n objects are in their natural place, then for two objects:
- p(2) =
- p(1) = 0
- p(0) =
With three objects, there are 3! (=6) arrangements.
- In one of these all three objects are in their natural place; so p(3) = .
- It's impossible to have exactly 2 objects in their natural place; so p(2) = 0
- There are three ways of choosing one of the objects to be in its natural place, and then just one way of arranging the remaining two so that neither is in its natural place; so p(1) =
- There are just two arrangements left from the six possible ones; so p(0) =
With four objects, there are 4! (=24) arrangements. Using similar arguments:
- p(4) =
- p(3) = 0
- For p(2), there are ways of choosing the two to be 'correct', and then just one way of making the remaining two 'incorrect'. So p(2) =
- For p(1), there are 4 ways of choosing the 'correct' one, and then 2 ways of arranging the remaining 3 where none is correct (see above, for 3 objects). So p(1) =
- This accounts for 1+6+8 = 15 of the 24 arrangements. So in the remaining 9, none is in the correct place. So p(0) = .
Now that we have established all the relevant individual probabilities, we can start to answer the question. I'll get you started:
(a) This will happen in one of three ways:
- 2 'correct' from 4 followed by 0 correct from 2. The probability of this is ... x ... (using the above results)
- 1 'correct' from 4, followed by 1 from 3. The probability of this is ...
- 0 correct from 4, followed by 2 from 4.
Work out these three probabilities and add them together.
(b) Use Bayes' Theorem. See here
Can you complete it now?