A box contains tickets marked 1,2,3...n. A ticket is drawn at random from the box. Then the ticket is then replaced in the box and a second ticket drawn at random. Find the probabilities of the following events. a) the first ticket drawn is a number 1 and the second is the number 2. b) the numbers on the two tickets are consecutive integers, meaning the first number drawn is one less than the second number drawn. c) the second number drawn is bigger than the first number drawn.

A) 1/n²

B) (n - 1)/n²

C) Probability = (1 - (1/n))/2

Step-by-step explanation:

A) We are told that a ticket is drawn at random from the box. Thereafter, the ticket is replaced in the box and a second ticket is now drawn at random.

This means the events of the first and second tickets are independent.

Probability of first ticket = 1/n

Probability of 2nd ticket = 1/n

Probability of both together = 1/n × 1/n = 1/n²

B) The number on the 2 tickets are consecutive integers.

This means number of cases will be (n - 1)

From a above, we see that for the two tickets the denominator is n²

Thus, in this case, the probability that both that the numbers on the two tickets are consecutive integers = (n - 1)/n²

C) if we assume that the first ticket is b, then the number of correct selections for the 2nd ticket is (n - b).

Thus, number of correct selections is;

(n_Σ_k=1) (n - b) = n² - (n(n - 1)/2)

Simplifying this gives;

(2n² - n² - n)/2

>> ½(n² - n)

There are n² selections in total, thus let's factorize n² out and then the remaining part in bracket will be the probability. Thus;

>> n²(1 - (1/n))/2

Probability = (1 - (1/n))/2

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8