ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: cara.mel on May 03, 2008, 12:15:14 pm

Title: Probability question
Post by: cara.mel on May 03, 2008, 12:15:14 pm: This is probably really easy: (heh, Pr(question easy) = 0.9)

Say I have 4 unbiased yadayada dice that each have 16 sides from 20-35, and I want the Probability the total is >= 100. (or if it floats your boat, once dice that I roll 4 times. :o)
Whats the answer :P

I can think of simplifying it to having them with sides 0-15 and wanting to get at least 20 from all 4.
The only way Ive learnt how is to draw a massive sample space which Im not doing tyvm

Oh and mean = 27.5, then I tried to work out standard deviation and got 27.8ish so I gave up
Title: Re: Probability question
Post by: Mao on May 03, 2008, 12:29:23 pm: i dont see a mathematical way of doing this (easily) from my knowledge, so, when i finish my maths assignment i'm going to ask VB =)
Title: Re: Probability question
Post by: Collin Li on May 03, 2008, 01:08:30 pm: Here is a problem that I would like solved:

A 16-sided die labelled 20 to 35 is rolled 4 times.

a) Find the probability that the sum of the numbers are greater than or equal to 100, but also so that the sum of the numbers for the first three rolls are less than 100.
(I would also like to know the probability of getting to 100 in 3 dice rolls)

b) Repeat part (a) with 5 dice rolls, and with the first 4 rolls being less than 100.
Title: Re: Probability question
Post by: Ahmad on May 03, 2008, 01:14:52 pm: I shall count the complement probability, although I won't do much explaining because not in the mood.

Want sum of coefficients with power less than 20 here:

$\frac{(1-x^{16})^4}{(1-x)^4}$

Equivalently powers less than 20 in:

$(1 - 4x^{16})(1 - x)^{-4} = (1-x)^{-4} - 4x^{16}(1-x)^{-4}$

Which is easily,

$\displaystyle \left(\sum_{i=0}^{19} (-1)^i \binom{-4}{i}\right) - 4\left(\sum_{i=0}^{3} (-1)^i \binom{-4}{i}\right) = 8715$

Therefore probability we seek is $\frac{56821}{65536}$
Title: Re: Probability question
Post by: Ahmad on May 03, 2008, 01:25:57 pm: Oh, and just to show that no calculator is needed, the last step is equivalently,

$\displaystyle \sum_{i=0}^{19} \binom{i+3}{3} - 4 \sum_{i=0}^{3} \binom{i+3}{3} = \binom{19+3+1}{3+1} - 4\binom{3+3+1}{4} = 8715$
Title: Re: Probability question
Post by: excal on May 28, 2008, 11:32:47 am: There's a long winded way of doing it by using the binomial c.d.f, given that they are all indepedent (such that your joint probability will be Pr(first) x Pr(second) .. x Pr(last)).

That is to find probability of Pr(X >= 25) x Pr(Y >= 25) ...

Then modifying X to 24, 23, 22, 21, 20 and adjusting for all combinations of Y, Z and A (2nd - 4th die).

Rinse and repeat for Y, Z and A.

Ahmad's method looks better, but I don't understand it.