(Bonus) Suppose 10 cards are generated which have the numbers 1,2,3,....10 on the front and a random permutation of these numbers on the backs of the cards. For example here is a possible set of 10 cards for one random permutation: Number on Front 1 2 3 4 6 8 9 10 Number on Back 3 2 5 6 8 7 10 9 These cards are placed face up on the table. They are then turned over in the following order: choose a random card, say card 3. Turn it face down and note that 5 is on the back. This means that the next card to be turned over is card 5 to find 4 on the back. Turn over 4, then 6, then 1 and note that this returns us to the original card (#3) that is already face down. This forms a loop of 5 cards, ( a loop of length 5), the number of cards that are, when the cycle is completed, face down. In this case this is cards numbered 3,5,4,6,1 .The remaining cards are still face up. Notice that if we had started with cards 7,7,9, or 10, we obtain a loop of length 2. If we start with card 2, the loop is of length 1. For the following questions, explain your answer in detail. (a) Again assume we have 10 cards numbered 1-10 on the front and back of the cards is formed from a random permutation of the integers 1,2,...10. Let X, be the length of the loop if we start by turning over card j (j = 1, 2, ...10). Find the probability distribution of Xj. (b) Suppose we find the value of Y which leads to the longest loop. Denote this Z = max{X1, X2, X3, ...X10}.Find P(Z = z) for z = 6, 7, 8, 9, 10. (c) Repeat (b) in the case that there are a total of n cards, labelled 1, 2, 3, ..., n. Find a numerical approximation for the value P(Z > > ) for large values of n