Probability Question (1 Viewer)

[axlenatore]

He, im just checking the answer from a past hsc question on probability
Its from the 1994 HSC 2 unit exam, question 10 (a) part (iii)

10 (a)
David has invented a games for one person. He throws two ordinary dice repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 9, David wins. If the sum is 7, Davud loses. If the sum is any other number, he continues to throw until it is 7 or 9
(iii) What is the probabilty that David wins on his first, second, or thrid throw? leave your anser in unsimplified form.

My answer was different to the one in the back, just checking whether the book is wrong or im just shit at maths

Thanks.

 

[dolbinau]

wins on first throw = 4/36

win on second throw. Now, 6 combinations (sum of 7) are removed because otherwise it means he will lose so the only way to win is if they are removed so probability = (26/36)(4/36)

win on third throw = (26/36)(26/36)(4/36)

So total probability equals the sum of them all.

I hate probability and it's probably wrong but I thought I'd have a go .

 

[Darrow]

For him to roll a nine he can get:
a 6 and 3
a 5 and 4
and the other way round
so out of the 36 combinations, there are 4 possibilities

ok, if he was to win on the first throw, it would be 4/36
Second throw we have to discredit all the values of rolling a 7
1 and 6
2 and 5
3 and 4 so 6/36 leaving 26/36
(26/36)(4/36)

Third throw is the same
(26/36)(26/36)(4/36)

Sounds good to me Dolbinau

 

[Timothy.Siu]

so final answer is 727/2916 or 0.249314128

 

[lyounamu]

He, im just checking the answer from a past hsc question on probability
Its from the 1994 HSC 2 unit exam, question 10 (a) part (iii)

10 (a)
David has invented a games for one person. He throws two ordinary dice repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 9, David wins. If the sum is 7, Davud loses. If the sum is any other number, he continues to throw until it is 7 or 9
(iii) What is the probabilty that David wins on his first, second, or thrid throw? leave your anser in unsimplified form.

My answer was different to the one in the back, just checking whether the book is wrong or im just shit at maths

Thanks.

10(a)
i)
...1 2 3 4 5 6
1 ...............x
2 ............x...
3 .........x.....o
4 ......x......o..
5....x.....o......
6 x....o..........

Chance of winning = 4/36 = 1/9

ii)
Second throw is needed when you get neither the "win" or "lose" = 26/36 = 13/18

iii)

David winning first throw = 1/9
David winning second throw = 13/18 . 1/9 (because he shouldn't have either won or lost in the first throw)
David winning thrid throw = 13/18 . 13/18 . 1/9 = 169/2916

Adding them altogether: 727/2916

iv) The series is formed as you see my above working out. It's an infinite sum since r = 13/18 < 1

S = a/(1-r) = 1/9/(1-13/18) = 2/5

 

[dolbinau]

Why is there a IV) part, isn't 727/2916 the final answer?

 

[lyounamu]

Why is there a IV) part, isn't 727/2916 the final answer?

Actually I did the whole question he outlined. That question has 4 parts to it and I just did that all. The answer for the iii) is 727/2916 as you said.

 

[axlenatore]

10(a)

iii)

David winning first throw = 1/9
David winning second throw = 13/18 . 1/9 (because he shouldn't have either won or lost in the first throw)
David winning thrid throw = 13/18 . 13/18 . 1/9 = 169/2916

Adding them altogether: 727/2916

Thanks, i got the same answer as you, obviously the answers in my book are wrong

 

[dolbinau]

Actually I did the whole question he outlined. That question has 4 parts to it and I just did that all. The answer for the iii) is 727/2916 as you said.

I see .

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What did they have in the actual answers? We could be all wrong.

 

[lyounamu]

I see .

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What did they have in the actual answers? We could be all wrong.

I don't know. As I don't have an access to the answer. I got the question from BOS.

 

[lyounamu]

Thanks, i got the same answer as you, obviously the answers in my book are wrong

What are the answers there then?

 

[axlenatore]

What are the answers there then?

This is from the 2008 edition Success one HSC Mathematics.

It had the following answers, there was working out but for (iii) i didnt really have working out for some reason.

(i) = 1/9

(ii) = 13/18

(iii) = (13/18)^2 x 1/9

(iv) = 2/5

I have found a few other errors in the book, but other than that its still good, the errors including wrong placing of a decimal place and incorrect multiplication. I just finished a past paper from it and in a superannuation question, the number from the gp was 1 decimal place too far to the right, causing the whole answer to be wrong. Still a worth wild investment