s********t
发帖数: 31 | 1 There are 10 red balls, 20 blue balls, and 30 green balls in a box.
You take them out one by one randomly.
Question:
What is the probability that when all red balls are taken out, there are
still at least 2 blue balls and 2 green balls in the box? | n****e
发帖数: 629 | 2 反过来拿呗
【在 s********t 的大作中提到】
: There are 10 red balls, 20 blue balls, and 30 green balls in a box.
: You take them out one by one randomly.
: Question:
: What is the probability that when all red balls are taken out, there are
: still at least 2 blue balls and 2 green balls in the box?
| s****n
发帖数: 1237 | 3 see Zhou's P79. But this one you need consider cases of
BBGG, BGBG, BGGB; GGBB, GBGB, GBBG instead of BG and GB
【在 s********t 的大作中提到】
: There are 10 red balls, 20 blue balls, and 30 green balls in a box.
: You take them out one by one randomly.
: Question:
: What is the probability that when all red balls are taken out, there are
: still at least 2 blue balls and 2 green balls in the box?
| s********t
发帖数: 31 | 4 The method in Zhou (usinng conditional probability) will work but the
process is a little complicated. For example, it is complicated if there
are at least 4 blue, 4 green balls before all red balls are taken?
Wonder if there is other method more elegant, e.g., using combinatorial
analysis?
【在 s****n 的大作中提到】
: see Zhou's P79. But this one you need consider cases of
: BBGG, BGBG, BGGB; GGBB, GBGB, GBBG instead of BG and GB
| p*****k
发帖数: 318 | 5 here is what i got for the prob of "at least N blue and N green
balls remaining when all the red balls are taken out" following
Zhou's method:
denote (r,b,g) as the number of balls of each color and t=r+b+g
is the total number of balls. apparently N <= b,g. then the prob:
C(b,N)*C(g,N)/C(t,N) * sum{of i from 0 to N-1}
C(N,i)*C(2N-1-i,i)/C(t-4,i)*[1/C(r+g-i,N-i)+1/C(r+b-i,N-i)]
where C(m,n)=m!/[n!(m-n)!] is the binomial coefficient.
r=10,b=20,g=30,N=2 gives 96311/266916 ~ 0.36 | f***a
发帖数: 329 | 6 denote (r,b,g) as the number of balls of each color
let i be the number of blue that will be left in the box
let j be the number of green that will be left in the box
then, i can be 2 to b, j can be 2 to g.
considering the last ball you pick out is red at the time that all red are
picked out, the balls still left in box are: i blue, j green.
because the last red ball is fixed (you pick it last), the others balls you
already picked out are: (r-1) red, (b-i) blue, (g-j) green.
The ways you can do | z****e
发帖数: 2024 | 7 bingo。
【在 p*****k 的大作中提到】
: here is what i got for the prob of "at least N blue and N green
: balls remaining when all the red balls are taken out" following
: Zhou's method:
: denote (r,b,g) as the number of balls of each color and t=r+b+g
: is the total number of balls. apparently N <= b,g. then the prob:
: C(b,N)*C(g,N)/C(t,N) * sum{of i from 0 to N-1}
: C(N,i)*C(2N-1-i,i)/C(t-4,i)*[1/C(r+g-i,N-i)+1/C(r+b-i,N-i)]
: where C(m,n)=m!/[n!(m-n)!] is the binomial coefficient.
: r=10,b=20,g=30,N=2 gives 96311/266916 ~ 0.36
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