math probability problem

>>8801442
get out of here with this shit. /sci/ is for pop psych personality and rationality quizzes.

start by finding the entire range of possible sums weighted by their probability of occurring

>>8801469

already did that, possibility for 11 on 4 dices is 104/1296, but question is how to find the possibility of sum 11 on 4 dices when you are throwing 5 dices.

With 5 dices there is 7776 possible outcomes, so... Counting them one by one would be more than insane.

>>8801442
You map out all the possible outcomes and then you circle the ones you've been asked for and add them up

>>8801476
then there's 6 ways for the 5th die to fall, given that the other 4 added up to 11

>>8801478
m8 it would be primary school method. it's 5 six sided dices. 7776 possibilities.

This is binomial distribution, and I tried to use Bernoulli process. It should provide, but I had to messed up with something.

>>8801482
Yep but how to implement this into calculations?

bumping for halp

>>8801516

lmao are you for real?

there are 104 × 6 possibilities - the times 6 is because the fifth dice can be everything

Use the principle of inclusion-exclusion.

>>8801552
Yes I am, because this is an error from your side.

104 x 6 is a 624

624 from 1296 possibilities for 4 dices is... 48%, so this is not true.When I throw 5 six sided dices probability for sum 11 on 4 dices can't be almost 50/50 :)


Okey 624 from 7776 is 8%... Same as 104 from 1296... So this is also not true m8.


You can't just multiply possibilities "because 5th dice".

Whenever you get a math problem written in text form the first you have to do is to write it in terms of mathematical formulas.

First define a probabilty space [math](\Omega, \mathscr{F}, P) [/math]. The obvious choice is to take
[math] \Omega = \{1,2,3,4,5,6 \}^5 [/math]
[math] \mathscr{F} = \mathscr{P}(\Omega) [/math]
[math] P(A) = \frac{|A|}{6^5} [/math]

Now define five random values [math]X_i: \Omega \to \mathbb{R} [/math] for [math]i=1,2,..,5[/math] whose values are the results of the dice rolls so
[math]X_i ((\omega_1, \omega_2, \omega_3, \omega_4, \omega_5)) = \omega_i [/math] for all [math](\omega_1, \omega_2, \omega_3, \omega_4, \omega_5) \in \Omega [/math].

Now for [math]i=1,2,..,5[/math] consider the events
[eqn]A_i = \left\{ \omega \in \Omega \middle| -X_i(\omega) + \sum_{k=1}^5 X_i(\omega) = 11 \right\} [/eqn]


With those definitions the probability you have to calculate is just
[eqn] P\left( \bigcup_{i=1}^5 A_i \right) [/eqn]
which can easily be calculated with >>8801558.

>>8801612
Thank you very much m8, I will try it now. :)