Try writing the expression with a summation sign. Also, try to figure out a way to write the powers in a cleaver way
This is a well known expression!
Hi to you all,
i am new at this forum and i am glad there is such.
Problem description
suppose i am playing two seperate games (independently!) with the following rules:
game #1- i have 4 empty cells in a row and i am trying to fill each cell (1 at the time) with red balls, the probability of filling each cell is p1 and (1-p1) for the cell to remain empty - -> only 4 attempts.
game #2- i have 4 empty cells in a row and i am trying to fill each cell (1 at the time) with black balls, the probability of filling each cell is p2 and (1-p2) for the cell to remain empty - -> only 4 attempts.
Question: what is the probabilty of having a total of n=1,2,3,...,8 filled cells? after a total of 4+4 attempts.
i have already solved the problem but i am looking for an elegant mathematical expression. right now my solution looks like:
p(n=0)=(1-p1)^4* (1-p2)^4
p(n=1)=4*(1-p1)^4*p2*(1-p2)^3+4*(1-p2)^4*p1*(1-p1)^3
p(n=2)=...etc.
Thanks in advanced.
Yaron B.S.
Try writing the expression with a summation sign. Also, try to figure out a way to write the powers in a cleaver way
This is a well known expression!
yaron (08-03-2014)
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