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These questions is from a book called Probability by Pitman.

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) I got this right right with $\frac{1}{n^2}$
b) the answer in the back of the book is $\frac{(n-1)}{n^2}$. I can see that the first term is $\frac{(n-1)}{n} \times Something$. From what I can see this term depends on what is chosen from the box on the first selection. I really don't know how to complete this.
c) it's the same kind of thing as b, the first selection would be $\frac{(n-1)}{n}$ but the probability for the second selection depends on what is actually selected in the firs selection. i.e. if 7 was chosen and n = 10, then it would be $\frac{3}{10}$ It's like $\frac{n-k}{n}$ where k was the first number chosen. But the actual answer in the back of the book is $\frac{(1-\frac{1}{n})}{2}$.

Any help out there?

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  • $\begingroup$ Ok, i think I worked it out. The second term can only be one number of n, because it's the consecutive number, therefore $\frac{1}{n}$ This would satisfy b. I will go away and look at c) after I get some pizza or someone can answer it if they like. Thanks. $\endgroup$
    – Bucephalus
    Jul 3, 2017 at 6:01
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    $\begingroup$ For part c, using a counting strategy. Count how many correct choices appear. For $k=1$, there are $n-1$ correct selections, for $k=2$, there are $n-2$ correct selections,... Probability is then the number of correct selections divided by the number of possible selections. $\endgroup$
    – Jonathan Davidson
    Jul 3, 2017 at 6:30
  • $\begingroup$ @JonathanDavidson, thankyou. I understand now, I have to use a series to get this answer. If you can put your answer up as an answer I can mark it off as answered. $\endgroup$
    – Bucephalus
    Jul 3, 2017 at 7:14

2 Answers 2

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a.) The events are independent. The probability of each event is $\frac{1}{n}$ so the joint probability is $\frac{1}{n^2}$

b.) There are $n-1$ cases where consecutive tickets are draw. Since there are $n^2$ possible ways to draw two tickets, the probability is $\frac{n-1}{n^2}$

c.) Suppose the first ticket picked is $k$, there are $n-k$ correct selections for the second ticket. The total number of correct selections is then $$\sum_{k=1}^{n}(n-k) = n^2-\frac{n(n-1)}{2} = \frac{n^2(1-1/n)}{2}$$ Since there are $n^2$ selections in total, the probability is $\frac{(1-1/n)}{2}$

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  • $\begingroup$ Cheers @JonathanDavidson $\endgroup$
    – Bucephalus
    Jul 3, 2017 at 7:30
  • $\begingroup$ No problem, i took the extra effort to correct you thinking on part b too. In these types of problems, counting is your friend. $\endgroup$
    – Jonathan Davidson
    Jul 3, 2017 at 7:32
  • $\begingroup$ ok deal, @Jonathan. $\endgroup$
    – Bucephalus
    Jul 3, 2017 at 7:34
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In (c) there are three possible outcomes: (i) the first number is bigger than the second, (ii) the first number is smaller than the second, (iii) the first number is equal to the second.

Surely (i) and (ii) have equal probability?

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  • $\begingroup$ Are you saying that on average, the value of the first selection will be $\frac{(n-1)}{2}$ and so the probability of the second term will be $\frac{(n-1)}{2}$. I still can't get the answer even with this. @LordSharktheUnknown $\endgroup$
    – Bucephalus
    Jul 3, 2017 at 6:51
  • $\begingroup$ Sorry, I mean $\frac{\frac{(n-1-1)}{2}}{n}$ $\endgroup$
    – Bucephalus
    Jul 3, 2017 at 6:59
  • $\begingroup$ Thanks for your help, @LordShark. I understand now. $\endgroup$
    – Bucephalus
    Jul 3, 2017 at 7:15

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