Probability of Peter's Village Visits

This question involves calculating probabilities related to the order in which Peter visits four different villages (A, B, C, and D). Since the visits are in a random order, we will use principles of permutations and relative order to find the required probabilities.

Total Possible Visit Orders

Peter visits 4 villages (A, B, C, D) in a random order. The total number of ways to arrange these 4 villages is given by the number of permutations of 4 distinct items, which is \(4!\).

So, there are 24 possible unique sequences in which Peter can visit the villages.

(i) Probability that Peter visits A before B

We need to find the probability that village A is visited before village B. When considering the relative order of any two distinct elements in a random permutation of a set, each element is equally likely to appear before the other.

• Consider only villages A and B. In any random arrangement of the 4 villages, either A comes before B, or B comes before A.
• Due to symmetry, the number of arrangements where A comes before B is exactly half of the total number of arrangements.
• Alternatively, the number of arrangements where B comes before A is also half of the total number of arrangements.

Thus, the probability that Peter visits A before B is:

Out of 24 total permutations, 12 will have A before B, and 12 will have B before A.

Therefore, \(P(\text{A before B}) = \frac{12}{24} = \frac{1}{2}\).

(ii) Probability that Peter visits A before B and B before C

Now we need to find the probability that Peter visits A before B, and B before C. This implies a specific relative order for villages A, B, and C as A → B → C.

• Consider the three specific villages A, B, and C. There are \(3!\) ways in which these three villages can be ordered relative to each other within any permutation of the four villages.
• The possible relative orders for A, B, and C are:
  • ABC (A before B, B before C)
  • ACB (A before C, C before B)
  • BAC (B before A, A before C)
  • BCA (B before C, C before A)
  • CAB (C before A, A before B)
  • CBA (C before B, B before A)
• Each of these \(3! = 6\) relative orders is equally likely in a random permutation.
• The condition "A before B and B before C" corresponds to only one of these 6 relative orders: ABC.

Therefore, the probability that Peter visits A before B and B before C is:

Out of the 24 total permutations, we can explicitly list them to verify. For example, some permutations where A → B → C are:

• ABCD
• ABDC
• DABC
• DABC (No, DABC has D before A. DABC is A B C in order with D at front.)
• ADBC
• CDAB (No, CDAB has C before D, then A before B. This does not satisfy B before C)

Let's consider the number of arrangements where A → B → C. Imagine A, B, C are fixed in that relative order. We effectively treat them as a block for ordering purposes, or simply consider their positions relative to each other. We are choosing 3 positions out of 4 for A, B, C in that specific order. The remaining position is for D. For example:

There are 4 such arrangements out of 24 total arrangements. So the probability is \( \frac{4}{24} = \frac{1}{6} \). This confirms our symmetry argument.

Summary of Probabilities

• Probability that Peter visits A before B: \( \frac{1}{2} \)
• Probability that Peter visits A before B and B before C: \( \frac{1}{6} \)

Comparing these results with the given options, the correct option is \( \frac{1}{2}, \frac{1}{6} \).