Probability word problem

[Guest]

I got as far as a .019% chance of a defective watch in 9046 watches, but the final line has got me struggling, anybody able to help with this!?

A wrist watch factory produced 9216 (=96*96) watches, among which 170 are defective. The 170 defective and 9046 non-defective watches were thoroughly mixed and randomly allocated and packaged into 96 boxes, of which each contains 96 watches. Find the probability that at least five boxes contains at least five defective watches

[Guest]

This is problem 112 from "115 Challenging and Interesting Probability Problems for High School and University Students and Teachers" there is a non-free version of the book with the solutions. Problems 93 to 115 in the book are all pretty similar just with different numbers (given 20% of the book is essentially the same it doesn't look like much effort was put into it). The book says
"a few very challenging problems need some special computing tools and programming codes"
I strongly suspect this is one of those problems, especially as it appears at the end of the book.

If you don't want to code you can come up with an approximate answer using the fact that [tex]p[/tex] = probability a watch is defective = 170/9216

Probability a chosen box has no defective watches = [tex](1-p)^{96}[/tex] (approx.)
Probability a chosen box has 1 defective watch = [tex]96p(1-p)^{95}[/tex] (approx.)
Probability a chosen box has 2 defective watches = [tex]\binom{96}{2}p^2(1-p)^{94}[/tex] (approx.)
Probability a chosen box has 3 defective watches = [tex]\binom{96}{3}p^3(1-p)^{93}[/tex] (approx.)
Probability a chosen box has 4 defective watches = [tex]\binom{96}{4}p^4(1-p)^{92}[/tex] (approx.)

So the probability that a box has 4 or less defective watches is
[tex]q=(1-p)^{96}+96p(1-p)^{95}+\binom{96}{2}p^2(1-p)^{94}+\binom{96}{3}p^3(1-p)^{93}+\binom{96}{4}p^4(1-p)^{92}[/tex] (approx.)

Probability no boxes have 5 or more defective watches = [tex]q^{96}[/tex] (approx.)
Probability 1 box has 5 or more defective watches = [tex]96(1-q)q^{95}[/tex] (approx.)
Probability 2 boxes have 5 or more defective watches = [tex]\binom{96}{2}(1-q)^2q^{94}[/tex] (approx.)
Probability 3 boxes have 5 or more defective watches = [tex]\binom{96}{3}(1-q)^3q^{93}[/tex] (approx.)
Probability 4 boxes have 5 or more defective watches = [tex]\binom{96}{4}(1-q)^4q^{92}[/tex] (approx.)

Probability 5 or more boxes have 5 or more defective watches is
[tex]1 - (q^{96}+96(1-q)q^{95}+\binom{96}{2}(1-q)^2q^{94}+\binom{96}{3}(1-q)^3q^{93}+\binom{96}{4}(1-q)^4q^{92})[/tex] (approx.)

Hope this helped,

R. Baber.

[Guest]

I thought about the problem a bit more and it is an interesting exercise to try to efficiently calculate the exact probability using a computer, if you're careful about how you do the calculation I think you should be able to do it in under an hour of computing time on a single core on most computers. I might have a go later.

Hope this helped,

R. Baber.

[Guest]

Thank you very much for helping me. The question was definitely not what I thought it was. I was expecting a simple probability question with a percentage or something.

[Guest]

My previous answer gives a probability of 21% (approx).
I tried to calculate the exact answer and got 16.9...% specifically:

44 387 430 519 681 406 295 114 509 634 454 316 566 047 622 806 066 811 250 896 285 531 591 015 808 217 134 722 745 714 516 107 966 793 960 722 662 019 907 628 599 183 968 328 790 922 014 013 879 996 303 510 626 805 302 756 797 964 927 235 240 336 481 219 008 626 334 589 844 664 269 020 185 707 634 506 623 492 047 678 571 079 362 416 950 536 799 551 667 993 048 554 316 664 932 935 564 900 756 028 341 427 328 348 004 368 119 571 963 565 477 875 754 833 895 351 215 709
/
262 149 237 919 847 927 389 582 635 450 283 030 798 055 134 735 046 582 405 670 366 263 757 868 888 908 063 856 082 037 501 943 567 014 466 957 875 029 163 938 223 690 284 770 880 381 230 373 893 641 822 348 759 809 812 801 935 388 265 950 474 597 148 435 136 602 404 185 967 420 578 959 169 548 069 418 780 730 807 255 450 101 033 542 934 671 826 696 183 636 544 336 266 577 251 676 279 780 163 282 728 608 125 838 196 849 235 391 855 743 568 163 571 254 918 638 620 617

Not sure if this is right though, I haven't really tested my program.

Hope this helped,

R. Baber.

[Guest]

I ran two simulations each consisting of 1 000 000 000 trials to estimate the probability. The first estimate came back as 16.9352...% the second came back as 16.9297...%. So my attempt at an exact answer is probably ok.

Hope this helped,

R. Baber.