You have a box of tacks for which it has been determined that if you toss them on the floor precisely 36% will land point up.
How many of these tacks could you toss to make the most likely result be that 10 of them will land point up.
Assuming that 36% is the expected number of pointed-up tacks, it's implied that each tack has probability 0.36 of coming up heads. Only an infinite-size box would assure the 36% ratio, but even a 100-tack box would have an expected number being 36.
As a start one can estimate 10/0.36 = 27.777... as being near the required answer.
The following program evaluates for given numbers of tacks tossed between 26 and 32, the probability of 8, 9, 10, 11 or 12 tacks landing upward:
10 for N=26 to 32
15 print N;
20 for Succ=8 to 12
30 Prob=combi(N,Succ)*0.36^Succ*0.64^(N-Succ)
40 print using(3,5),Prob;
50 next Succ
60 print
70 next N
It finds:
N 8 9 10 11 12
26 0.14303 0.16091 0.15387 0.12589 0.08852
27 0.13008 0.15447 0.15640 0.13596 0.10197
28 0.11655 0.14569 0.15570 0.14332 0.11421
29 0.10301 0.13520 0.15210 0.14778 0.12469
30 0.08990 0.12361 0.14601 0.14933 0.13300
31 0.07755 0.11147 0.13795 0.14814 0.13888
32 0.06617 0.09926 0.12842 0.14447 0.14221
So, if 27, 28 or 29 tacks were tossed, 10 of them pointing up would be the most likely outcome, at probabilities of 0.15640, 0.15570 and 0.15210 respectively. The 27 count maximizes this probability, but all three values satisfy the requirement that 10 be the most likely outcome.
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Posted by Charlie
on 2011-12-01 13:32:42 |
computer-assisted solution
