Whats the probability of me failing my exam tomorrow?

qwertykeyboards · 18-12-2008 09:20PM #1

pretty high considering i cant do these questions or most others.

any help muchly appreciated!

Thomas_S_Hunterson · 18-12-2008 09:25PM

First question, just count up the combinations.

For the second, use the conditional probability law /edt: or Bayes
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MathsManiac · 18-12-2008 09:35PM

1 - P(you pass) - P(you don't do an exam tomorrow)

Good luck!

qwertykeyboards · 18-12-2008 10:00PM

thanks for your help. for the first i got 25? and second, which is A and which is B?

Thomas_S_Hunterson · 18-12-2008 10:03PM

For the second you have to apply the formula a couple of times to different things to get the information you need.
solution

This STAT20010 btw?

qwertykeyboards · 18-12-2008 10:11PM

thanks for the help again. no its a probability module in dcu. did u write up that solution yourself or is there a site without solutions to a load of similar problems?

Thomas_S_Hunterson · 18-12-2008 10:16PM

No I wrote it up, I don't think your answer is right to the first one.

qwertykeyboards · 18-12-2008 10:23PM

ah fair play. no its defo not right. you really have to read every single word carefully with these poxy questions. i worked out how many different totals u can get. i'll give it another bash. thanks.

Grudaire · 18-12-2008 11:08PM

qwertykeyboards wrote: »

ah fair play. no its defo not right. you really have to read every single word carefully with these poxy questions. i worked out how many different totals u can get. i'll give it another bash. thanks.

Ah I'm in that class! if it helps I emailed old Menkins to explain that first question

it means who often any of the numbers 1 to 6 happens in the five throws.

Ans that I got was:

462

red_fox · 19-12-2008 01:53AM

There are two (main) ways to view the first. 'Choosing' 5 objects from 6 allowing repetition (aka with replacement) or placing 5 objects between 6 catagories.

i.e. solutions to x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 5 where x_i >= 0

in you case x_i would be the number of dice that gace value i.

In general x_1 + x_2 + ... + x_k = n would have solution (n+k-1)C(n), that can be shown with a nice argument using bars and objects but you'll find that in any textbook on the subject.

LeixlipRed · 19-12-2008 10:52AM

You have Menkens? I pity you

Fremen · 22-12-2008 09:00PM

Looks like I dodged a bullet lexlip

LeixlipRed · 23-12-2008 02:01AM

Haha, yeh. A mate of mine is currently doing a Phd with said chap. I feel sorry for him

Whats the probability of me failing my exam tomorrow?

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