Although some of the things in this problem aren't true in real life just assume they are in the question.

When growing peanuts the following happens:

for every 2 single chambered peanuts there will be one double chambered peanut

for every 2 double chambered peanuts there will be one triple chambered peanut

for every 2 triple chambered peanuts there will be one quadruple chambered peanut.

When a company packages 1000 peanuts in one bag they take peanuts randomly from a giant bin that contins all the peanuts grown. What are the odds that there will be 1000 individual nuts?

(In reply to Slight Correction by Charlie)

As a check, here's the program logic; it goes through all sum of compartments = 1000 and counts only those adding to 577:
80 P1=15:P2=15/2:P3=15/4:P4=15/8
90 point 25
100 T=1000
110 for Quad=0 to int(1000/4)
120 T2=T-Quad*4
130 for Trip=0 to int(T2/3)
140 T3=T2-Trip*3
150 for Dbl=0 to int(T3/2)
160 Sngl=T3-Dbl*2
161 if Quad+Trip+Dbl+Sngl=577 then
162 :W=combi(577,Quad)*combi(577-Quad,Trip)*combi(577-Quad-T
rip,Dbl)
163 :P=P1^Quad*P2^Trip*P3^Dbl*P4^Sngl
164 :Tp=Tp+W/P
170 next
180 next
190 next
200 print Tp

Posted by Charlie on 2003-02-24 09:13:00