probability question - wrong answer? (1 Viewer)

[tooheyz]

hey guys im just doing the 1988 past paper - question 9 part b(ii)

i dont know how they got the answer

for those who dont have the paper heres the question...

b (i) a die whose faces are numbered 1-6 is tossed twice. the sum S of the numbers which appear uppermost on the die is calculated. find the probability that S is greater than 8

( i got this answer, it was 5/18)

(ii) what i cant do... it is known that a 4 appears on the die at least once in 2 throws. find the prob that S is greater than 8

answer is 4/11

thanks to those who can help! cheers.

 

[Winston]

lol i did that question well

you see your first part of the answer helped you you found the numbers that's sum were greater than 8

this sum was formed by two numbers


so you go back and look at your table and check which set of numbers sum had a '4' in it to make it greater than 8


example, 5 and 4

= 9 it has a 4 in it and it's greater than 8

so that's one just look at the table for all the numbers greater than 8 which consisted a 4 to produce that sum.


I also agree that the 1980's past papers questions had the fucked up wording ques ever!

 

[dnt stop runnin]

yeh i did that question too..........i have no clue how they got 4/11.. neither do my teachers......im guessing its a wrong answer coz where they got 11 from beats me...........anyone knows, let me know 2

 

[tooheyz]

yeah man i got 4, but how on earth they got 4/11?

i might sound stupid i know...

 

[tooheyz]

Originally posted by dnt stop runnin
yeh i did that question too..........i have no clue how they got 4/11 neither do my teachers......im guessing its a wrong answer coz where they got 11 from beats me...........anyone knows, let me know 2

ah alright man - good to know that somebody else had the same problem - i thought i was loosing it

ah so its the wrong answer... it should be 4/10

coz theres 10 thingys... hmm...

alright i wont stress no more.

 

[Winston]

Yep the answer is incorrect

 

[Jumbo Cactuar]

yeah the P(S>8) = 10 x 1/36 = 5/18
and P(S>8 1d=4) = 4 x 1/36 = 1/9

 

[Winston]

Re: Re: probability question - wrong answer?

Originally posted by symo
4,5
5,4
4,6
6,4
4,4
4,4 (4 appears at least once, therefore can happen twice)

1/36 x 6 = 1/6 ?????

it has to be greater than 8!
therefore a sum of 8 is not valid

 

[HSC :(]

the answer is right

hai the answer 4/11 is correct

now in the question they specify that a four appears on the dice at least once in 2 throws, which means u gotta look at all the sample space that hav four in it

drawin a table here would b most suitable

1) 11 21 31 41 51 61
2) 21 22 23 24 25 26
3) 31 32 33 34 35 36
4) 41 42 43 44 45 46
5) 51 52 53 54 55 56
6) 61 62 63 64 65 66
1) 2) 3) 4) 5) 6)

so what u gotta do is the sample space here is 41 42 43 44 45 46 41 42 43 45 46

now u hav 11 possibilities, and from those 11, u gotta see which has a sum of 8, and u see 45, 46, 45. and 46 have that

so, its 4/11

 

[Winston]

Re: the answer is right

Originally posted by HSC
hai the answer 4/11 is correct

now in the question they specify that a four appears on the dice at least once in 2 throws, which means u gotta look at all the sample space that hav four in it

drawin a table here would b most suitable

1) 11 21 31 41 51 61
2) 21 22 23 24 25 26
3) 31 32 33 34 35 36
4) 41 42 43 44 45 46
5) 51 52 53 54 55 56
6) 61 62 63 64 65 66
1) 2) 3) 4) 5) 6)

so what u gotta do is the sample space here is 41 42 43 44 45 46 41 42 43 45 46

now u hav 11 possibilities, and from those 11, u gotta see which has a sum of 8, and u see 45, 46, 45. and 46 have that

so, its 4/11

it's a sum GREATER than 8

it can't be 8 it must be greater than it, so in this case anything from 9 and above is included but not 8 itself

 

[HSC :(]

UMM

WELL if u ACTUALLY added the sample spaces i hav included, like 45, 46 etc youd know the sum is greater then 8.. . 4+5= WAIT FOR IT 9!!!

 

[Winston]

Re: UMM

Originally posted by HSC
WELL if u ACTUALLY added the sample spaces i hav included, like 45, 46 etc youd know the sum is greater then 8.. . 4+5= WAIT FOR IT 9!!!

but... it's still not a total of 11

it's 10!

 

[slyball]

how are you getting ten possibilities?

humbo cactuar: and P(S>8 1d=4) = 4 x 1/36 = 1/9

i don't think that is right because you have to take into account that one of the die thrown is a 4. and 36 possibilities come from dice without that restriction.

combinations of dice that have a 4 in it:
1) 4+1
2) 4+2
3) 4+3
4) 4+4
5) 4+5
6) 4+6
7) 1+4
8) 2+4
9) 3+4
10) 5+4
11) 6+4

eleven possibilities.
nb. i counted 4+4 only once.
now, the only combinations with sum greater than 8 are 5), 6), 10) and 11). 4 chances in 11 possibilities with a 4 as one of the die with sum that is greater than 8.

4/11.

 

[Winston]

Originally posted by slyball
how are you getting ten possibilities?



i don't think that is right because you have to take into account that one of the die thrown is a 4. and 36 possibilities come from dice without that restriction.

combinations of dice that have a 4 in it:
1) 4+1
2) 4+2
3) 4+3
4) 4+4
5) 4+5
6) 4+6
7) 1+4
8) 2+4
9) 3+4
10) 5+4
11) 6+4

eleven possibilities.
nb. i counted 4+4 only once.
now, the only combinations with sum greater than 8 are 5), 6), 10) and 11). 4 chances in 11 possibilities with a 4 as one of the die with sum that is greater than 8.

4/11.

Oh i see, i think the question was rather decieving a bit, i just realised it was thhe ones that had a four in it silly me.

 

[kazerati]

yeh, see, thats wat the hsc does to you!!!!

i just read a whole PAGE of u'z all confused about a simple question - confused for the simple reason of you not being able to count...

i must ask - 'where is the wisdom we have lost in knowledge? where is the knowledge we have lost in information?'

sigh.. to think that in a matter of weeks this whole messy organisation of confusion will have finished its nasty business for yet another year...

i just want to be able to count again!! to add all those simple numbers like 4+5 without my calculator... sigh..