10 thoughts on “Oxford PAT 2010, Questions 9 and 10”

I think the answer should be 5/11
If you draw a tree diagram you can see all the possible outcomes are
(1+0) , (2+0) , (3+0), (4+0) , (5+0), (6+1) , (6+2), (6+3), (6+4), (6+5), (6+6)
there is no (6+0) as if he gets a 6 he has to throw another dice
so i think the total outcomes are 11. Please correct me if i am wrong

There are 11 outcomes, but they are not equally likely. The first five outcomes each happen six times as often as the outcomes where a six is thrown. Have a look at the answer to Justin below — if you try it out in a spreadsheet, you end up getting 15/36.

I believe you have included the probability of gaining two 6s. If he rolled a second six he will have to role more than twice. In the question It States he will role no more than two dice. By this reasoning I got the answer 7/18. Do you have any thoughts on this.

Tom

Keep up the good work with the site. It is really helpful.

So your reading of the question is essentially this: “if at any point the player throws a six, he/she then throws the die again, but the player can’t throw the die more than twice, therefore if she/he throws two sixes the whole throw is invalid”.

You’re right that this reading would give you an answer of 7/18, because the score of 12 would be invalid and therefore there would be 14 outcomes out of 36 that give an event number.

But I don’t think your reading is the right one. The question says “… initially throws a single die, and … . If the die shows a six, …” so I am pretty sure what it is saying is that the check on the value of the die only happens once, at the initial throw.

Thanks for this great site, it’s helped me out a lot. Could you please elaborate the case for question 10? I got the same answer as Justin. My argument is that if we list out all the possibilities:

1
2
3
4
5
7
8
9
10
11
12

Since you can never obtain a 6, the probability is 5/11.
The way we get your answer is by including 6 to obtain a 5/12 probability (=15/36)

Thanks for your comment. I’m really pleased the site has been useful.

If we consider your argument, it includes an unstated assumption that all the outcomes you have listed are equally probable. But they are not. The probability of 1, 2, 3, 4 and 5 is 1/6; and the probability of all the other outcomes put together (i.e. 7, 8, 9, 10, 11, 12) is 1/6 (i.e. the probability of initially throwing a six). Each of the numbers from 7 – 12 is equally probable: this means that the probability of each of 7 – 12 is 1/36. So when you add up the probabilities of the even numbers you get 15/36, which is 5/12.

These probability questions are hard to think about, and sometimes it’s worth having a play with a tool like Excel, as I described in the answer to Justin.

Hi Anna. Yes you are right, thanks for pointing that out. I guess I was thinking that, for the first term, n=0. But notice that the only thing that matters in the question is the difference between terms space 3 terms apart. This is why the answer doesn’t depend on whether we write a+nd or a+(n-1)d.

I don’t believe this solution is correct. Saying that the probability of getting a 8,10, or 12 is 1/36 is implying that 100% of the time there will be two thrower dice, which is not true. The dice will only be thrower twice when the player rolls a 6 n his first trial. Thus, I believe the answer is 5/11

I don’t think your argument is right. Here is another way of looking at it. Suppose we play this game some very large number of times, call it N. In N/6 games, we will throw a 2 and then stop. In N/6 games we will throw a 4 and then stop. In N/6 games we will throw a 6 and throw again. Let’s suppose that we throw again in M games: in M/6 games we will throw a 2, giving a total of 8, and similarly for 4 (total of 10) and 6 (total of 12). But M = N/6, therefore we will get a total of 8 in N/36 games.

If you are lucky enough to have a copy of Excel, it can sometimes help to do a little probability experiment. In this example, you can set up a row 1 in Excel like this:
A1: =RANDBETWEEN(1,6)
B1: =IF(A1=6,6+RANDBETWEEN(1,6),A1)
C1: =IF(MOD(B1,2)=0,1,0)
then copy this row and paste it into a very large number of rows below. Now if you sum the column C and divide by the number of rows, you will get the probability of getting an even number. I have just done this and with 30000 rows I make the probability 0.419, which equals about 15.08/36.

I hope this helps, and very best of luck with the exam!

I think the answer should be 5/11

If you draw a tree diagram you can see all the possible outcomes are

(1+0) , (2+0) , (3+0), (4+0) , (5+0), (6+1) , (6+2), (6+3), (6+4), (6+5), (6+6)

there is no (6+0) as if he gets a 6 he has to throw another dice

so i think the total outcomes are 11. Please correct me if i am wrong

There are 11 outcomes, but they are not equally likely. The first five outcomes each happen six times as often as the outcomes where a six is thrown. Have a look at the answer to Justin below — if you try it out in a spreadsheet, you end up getting 15/36.

I believe you have included the probability of gaining two 6s. If he rolled a second six he will have to role more than twice. In the question It States he will role no more than two dice. By this reasoning I got the answer 7/18. Do you have any thoughts on this.

Tom

Keep up the good work with the site. It is really helpful.

So your reading of the question is essentially this: “if at any point the player throws a six, he/she then throws the die again, but the player can’t throw the die more than twice, therefore if she/he throws two sixes the whole throw is invalid”.

You’re right that this reading would give you an answer of 7/18, because the score of 12 would be invalid and therefore there would be 14 outcomes out of 36 that give an event number.

But I don’t think your reading is the right one. The question says “… initially throws a single die, and … . If the die shows a six, …” so I am pretty sure what it is saying is that the check on the value of the die only happens once, at the initial throw.

Thanks for this great site, it’s helped me out a lot. Could you please elaborate the case for question 10? I got the same answer as Justin. My argument is that if we list out all the possibilities:

1

2

3

4

5

7

8

9

10

11

12

Since you can never obtain a 6, the probability is 5/11.

The way we get your answer is by including 6 to obtain a 5/12 probability (=15/36)

Cheers

Hi Rishi

Thanks for your comment. I’m really pleased the site has been useful.

If we consider your argument, it includes an unstated assumption that all the outcomes you have listed are equally probable. But they are not. The probability of 1, 2, 3, 4 and 5 is 1/6; and the probability of all the other outcomes put together (i.e. 7, 8, 9, 10, 11, 12) is 1/6 (i.e. the probability of initially throwing a six). Each of the numbers from 7 – 12 is equally probable: this means that the probability of each of 7 – 12 is 1/36. So when you add up the probabilities of the even numbers you get 15/36, which is 5/12.

These probability questions are hard to think about, and sometimes it’s worth having a play with a tool like Excel, as I described in the answer to Justin.

Isn’t the nth term of the progression = a + (n – 1)d? But I think you got the correct answer anyway, I checked with another version of the answer.

Hi Anna. Yes you are right, thanks for pointing that out. I guess I was thinking that, for the first term, n=0. But notice that the only thing that matters in the question is the difference between terms space 3 terms apart. This is why the answer doesn’t depend on whether we write a+nd or a+(n-1)d.

I don’t believe this solution is correct. Saying that the probability of getting a 8,10, or 12 is 1/36 is implying that 100% of the time there will be two thrower dice, which is not true. The dice will only be thrower twice when the player rolls a 6 n his first trial. Thus, I believe the answer is 5/11

Hi Justin — thanks very much for your comment.

I don’t think your argument is right. Here is another way of looking at it. Suppose we play this game some very large number of times, call it N. In N/6 games, we will throw a 2 and then stop. In N/6 games we will throw a 4 and then stop. In N/6 games we will throw a 6 and throw again. Let’s suppose that we throw again in M games: in M/6 games we will throw a 2, giving a total of 8, and similarly for 4 (total of 10) and 6 (total of 12). But M = N/6, therefore we will get a total of 8 in N/36 games.

If you are lucky enough to have a copy of Excel, it can sometimes help to do a little probability experiment. In this example, you can set up a row 1 in Excel like this:

A1: =RANDBETWEEN(1,6)

B1: =IF(A1=6,6+RANDBETWEEN(1,6),A1)

C1: =IF(MOD(B1,2)=0,1,0)

then copy this row and paste it into a very large number of rows below. Now if you sum the column C and divide by the number of rows, you will get the probability of getting an even number. I have just done this and with 30000 rows I make the probability 0.419, which equals about 15.08/36.

I hope this helps, and very best of luck with the exam!