Question

What is the probability of these events when we randomly select a permutation of {1,2, ..., n} where n≥4? a) 1 precedes 2 c) 1 immediately precedes 2 d) n precedes 1 and n−1 precedes 2

Answer 1

Answer:

a) 1/2

b) 1/n

c) 1/4

Step-by-step explanation:

a) For each permutation, either 1 precedes 2 or 2 precedes 1. For each permutation in which 1 precedes 2, we can swap 1 and 2 to obtain a permutation in which 2 preceds 1. Thus, half of the total permutations will involve in 1 preceding 2, hence, the probability for a permutation having 1 before 2 is 1/2.

c) If 2 is at the start of the permutation, then it is impossible for 1 to be before 2. If that is not the case, then 1 has a probability of 1/n-1 to be exactly in the position before 2. We can divide in 2 cases using the theorem of total probability,

P( 1 immediately preceds 2) = P (1 immediately precedes 2 | 2 is at position 1) * P(2 is at position 1) + P(1 immediately precedes 2 | 2 is not at position 1) * P(2 is not at position 1) = 0 * 1/n + (1/n-1)*(n-1/n) = 1/n.

d) We can divide the total of permutations in 4 different groups with equal cardinality:

• Those in which n precedes 1 and n-1 precedes 2
• those in which n precedes 1 and 2 precedes n-1
• those in which 1 precedes n and n-1 precedes 2
• those in which 1 precedes n and 2 precedes n-1

All this groups have equal cardinality because we can obtain any element from one group from another by making a permutations between 1 and n and/or 2 and n-1.

This means that the total amount of favourable cases (elements of the first group) are a quarter of the total, hence, the probability of the event is 1/4.