5d) Probability (1 Viewer)

[casomerville]

I got 81 ways for part i.

0.0144 for part ii.

31/81 for part iii.


Any takers?

 

[gamecw]

(i)81
(ii)0.0144
(iii)not attempted

 

[casomerville]

I think you had to work out all the different cases.

Eg: 4W, 3W 1L, 2W 2L, etc, etc.
Then work out how many of those arrangements yeiled more points for the home team. Which i think was 31.

Hence probablilty = 31/81 (i hope)

 

[zeek]

for the first one i said you can choose from 3 elements (W,L,D) .: 3 ways and ordering these is 4! so the total was 3.4!

 

[robmeister]

I got 81 ways for part i.

0.0144 for part ii.

31/81 for part iii.


Any takers?

I got exactly the same we are awesome, but i think we are wrong, that would be like a 2 unit questions difficulty. I even pondered this question while i was urinating during the test, i pissed for about 2 mins i was soo busting, but yeah couldnt think of any other way to do it.

 

[Gecko888]

For the last one I got like 0.344 - I don't think you can just say some arrangements yield more points for the home team, because the arrangements all have different probs, for example wwdd has a different prob than wwld. I wrote down the different selectinos, calculated the prob and multiplied by the number of permutations for each one

 

[XcarvengerX]

(i) How do you guys get 81? I did try to calculate by writing them out, and I didn't think it is 81.
(ii) Everyone got this.
(iii) You just find all the combination where the Home team win, THEN calculate like you did in part (ii) for each of these combination. I think it is around 0.07.

 

[Bricnic]

For the last one I got like 0.344 - I don't think you can just say some arrangements yield more points for the home team, because the arrangements all have different probs, for example wwdd has a different prob than wwld. I wrote down the different selectinos, calculated the prob and multiplied by the number of permutations for each one

I think you're right there Gecko. You could probably answer this question similarly to a binomial probability question, but as a trinomial instead.. or maybe it could be simplified to a binomial somehow. I didn't bother with part iii anyway, just answered parts i and ii, got 81 ways because its 3⁴ and 0.0144 for ii.

 

[Bricnic]

(i) How do you guys get 81? I did try to calculate by writing them out, and I didn't think it is 81.
(ii) Everyone got this.
(iii) You just find all the combination where the Home team win, THEN calculate like you did in part (ii) for each of these combination. I think it is around 0.07.

iii) 0.07? That seems awfully low...
Obviously the probability will be less than 0.5, because the home team will have just as much chance of winning as the away team. If the chance of the home team getting more points was just 0.07, then the chance of the away team getting more points must have been 0.07 as well. This would give a probability of 0.86 of both teams getting exactly the same number of points, which seems quite unlikely.

 

[ViRtUaL]

(i) 81
(ii) 9/625
(iii) 167/625 <-- i think this is wrong since two ppl have agreed on 0.344 and a mate of mine got that as well

 

[ravdawg]

(i) 81
(ii) 9/625
(iii) 167/625 <-- i think this is wrong since two ppl have agreed on 0.344 and a mate of mine got that as well

i got 0.344, far out though made worst mistake on part i did 4^3 instead of 3^4, biggest idiot ever!!

i didn't have enough time for this test!!!

 

[jon0_o]

yeah i got 31/81 but im pretty sure it's wrong now ... i think it's 43/125 now. I got the probability of an overall draw .. then the probability of not a draw and then that divided by 2 since both teams have an equal chance of winning. but i didn't take into account the fact that certain groupings lyk DDDD and DDWL have a difference chance of happening

 

[speedie]

its definitely 0.344, but i wrote down a diff answer :S

 

[XcarvengerX]

I think I forgot few combinations.

The chance of draw is 0.6, so if the home team has 0.344 chance to win, the away team only has 0.056 chance to win which is awfully low. Or is it 0.1296 the chance of draw, which means the away team has higher chance of winning? I don't know, did you use binomial distribution or simple counting, permutation techniques?

What's this about 81, 125? How do you get part (i)?

 

[milton]

part (i)
3 different outcomes for each of the 4 boards
i.e. 3^4 = 81

part (ii)
0.2 * 0.6 * 0.2 * 0.6 = 0.0144

part (iii)
initally i thought about listing the 6 different ways the Home team could win but that was too tedious
alternatively:

due to symmetry, Pr(Home scores more points) = Pr(Away scores more points)
and Pr(Home scores more points) + Pr(Away scores more points) + Pr(same number of poitns for both teams) = 1.

Consider Pr(same num of points)
Each team will score the same number of wins as losses, so the possibilties are:
2 wins, 2 losses => 4C2 * 0.2^4
1 win, 1 loss, 2 draws => 4C2 * 2 * 0.2 * 0.2 * 0.6 * 0.6
4 draws => 0.6^4
sum = 0.312

thus, Pr(Home scores more) = 0.344

 

[mrxt]

For part iii), I saw it as:

Total points awarded to all teams = 4

therefore, Home must score >2.

Possibilities:
1 Win, 3 Draws = 0.2 * (0.6)^3
2 Wins, 1 Draw = (0.2)^2 * (0.6)^2
3 Wins [Includes 4 wins] = (0.2)^3
Total Possibilities = 0.0656

Not sure if this is right though...

 

[crazedmo2003]

I did it in the same way as Milton and Bookie



BUT I forgot the choose coefficient (silly me)