Bu yaz\u0131da k\u0131sa bir hik\u00e2yesi olan ilgin\u00e7 bir olas\u0131l\u0131k sorusunu ele alaca\u011f\u0131z.<\/p>\n
Bir \u00fcniversitenin mezuniyet t\u00f6reninde \u00fcniversite y\u00f6netimi olas\u0131 protestolar\u0131 engellemek amac\u0131yla her \u00f6\u011frenciye bir numara vererek herkesin kendi numaras\u0131na ait koltu\u011fa oturmas\u0131 kural\u0131n\u0131 koyar. 500 ki\u015filik konferans salonunun giri\u015f kap\u0131s\u0131nda 500 \u00f6\u011frenci numaralar\u0131na g\u00f6re s\u0131ralan\u0131r, ilk s\u0131rada 1, ikinci s\u0131rada 2 numaral\u0131 \u00f6\u011frenci ve di\u011fer \u00f6\u011frenciler de bu \u015fekilde s\u0131ralanarak en sonda 500 numaral\u0131 \u00f6\u011frencinin bulundu\u011fu uzun bir kuyruk olu\u015fur. \u00d6\u011frencilerden salona s\u0131ra numaralar\u0131na g\u00f6re girmeleri istenmi\u015ftir. \u0130lk s\u0131radaki 1 numaral\u0131 \u00f6\u011frenci 1 numaral\u0131, 2 numaral\u0131 \u00f6\u011frenci 2 numaral\u0131 koltu\u011fa ve bu \u015fekilde her \u00f6\u011frenci \u00f6n\u00fcndeki \u00f6\u011frencinin kendi koltu\u011funa oturmas\u0131ndan sonra salona girerek, kendi numaras\u0131na ait olan koltu\u011fa oturmal\u0131 duyurusu yap\u0131l\u0131r.<\/p>\n
Salona girmek i\u00e7in bekleyen \u00f6\u011frenciler bu ak\u0131l d\u0131\u015f\u0131 uygulamaya polisiye tedbirler y\u00fcz\u00fcnden tepki vermekten korkarlar, ama bu s\u0131rada 1 numaral\u0131 \u00f6\u011frenci 2 numaral\u0131 \u00f6\u011frencinin kula\u011f\u0131na \u015fu s\u00f6zleri f\u0131s\u0131ldar: \u201cKoltuk numaralar\u0131na bakmaks\u0131z\u0131n rastgele bir koltu\u011fa oturarak k\u00fc\u00e7\u00fck bir oyun oynayaca\u011f\u0131m. Belki de kendi koltu\u011fuma, yani 1 numaral\u0131 olana otururum, bilemiyorum. B\u00f6ylece y\u00f6netici amcalar\u0131n isteklerinin ger\u00e7ekle\u015fme olas\u0131l\u0131\u011f\u0131 500 de 1 olacak\u201d. Bu muzip \u00f6\u011frenci dedi\u011fini yaparak rastgele bir koltu\u011fa oturur ve sonra gelen \u00f6\u011frenciler ceza alma korkusuyla itaatk\u00e2r davran\u0131p, kendi numaralar\u0131na ait koltuk bo\u015fsa o koltu\u011fa, bo\u015f de\u011filse rastgele bir koltu\u011fa otururlar.<\/p>\n
Kuyru\u011fun sonuna, son s\u0131radaki 500\u2019\u00fcnc\u00fc \u00f6\u011frenciye gelirsek\u2026 Matematik b\u00f6l\u00fcm\u00fcn\u00fcn en parlak \u00f6\u011frencilerinden biri olan bu mezun, 1 numaral\u0131 \u00f6\u011frencinin verdi\u011fi karar\u0131n ve sonra gelen \u00f6\u011frencilerin kendi numaralar\u0131na ait koltuk bo\u015fsa o koltu\u011fa, bo\u015f de\u011filse rastgele bir koltu\u011fa oturduklar\u0131n\u0131n haberini alm\u0131\u015ft\u0131r. Salona girmeyi beklerken kafas\u0131na \u015fu soru tak\u0131l\u0131r: Acaba benim kendi koltu\u011fuma oturma olas\u0131l\u0131\u011f\u0131m ka\u00e7?<\/p>\n
Hik\u00e2yele\u015ftirdi\u011fimiz bu ilgin\u00e7 olas\u0131l\u0131k problemini \u00f6zetleyerek \u015f\u00f6yle soral\u0131m: 500 \u00f6\u011frenciden ilk \u00f6\u011frenci rastgele bir koltu\u011fa oturuyor. Sonra s\u0131ras\u0131yla gelen \u00f6\u011frenciler e\u011fer kendi koltuklar\u0131 bo\u015f ise kendi yerlerine, bo\u015f de\u011filse rastgele bir yere otururlarsa 500\u2019\u00fcnc\u00fc \u00f6\u011frencinin kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r?<\/p>\n
Zor gibi g\u00f6r\u00fcnen bu problemin olduk\u00e7a basit bir \u00e7\u00f6z\u00fcm\u00fc var: Son \u00f6\u011frencinin salona girmesinden \u00f6nce sadece iki koltuktan biri bo\u015f olabilir: 1\u2019inci veya 500\u2019\u00fcnc\u00fc koltuklar, \u00e7\u00fcnk\u00fc bu koltuklar d\u0131\u015f\u0131ndaki yerlere ya sahipleri ya da ba\u015fkalar\u0131 oturmu\u015f olacak. O halde son \u00f6\u011frenci ya kendi koltu\u011funa ya da 1\u2019inci koltu\u011fa oturabilir, dolay\u0131s\u0131yla istenen olas\u0131l\u0131k 1\/2\u2019dir.<\/p>\n
Bu cevaba anla\u015f\u0131lmad\u0131\u011f\u0131 yolunda hakl\u0131 itirazlar gelebilir. Yukar\u0131daki \u00e7\u00f6z\u00fcm\u00fc daha a\u00e7\u0131k hale getirebilmek i\u00e7in \u00f6\u011frenci say\u0131s\u0131n\u0131 k\u00fc\u00e7\u00fcltelim. \u00d6rne\u011fin A<\/em>, B<\/em>, C<\/em>, D <\/em>gibi d\u00f6rt \u00f6\u011frencinin koltuk numaralar\u0131 s\u0131ras\u0131yla 1, 2, 3, 4 olsun ve salona A<\/em>, B<\/em>. C<\/em>, D <\/em>s\u0131ras\u0131yla girsinler.<\/p>\n A <\/em>= 1 ise yani A<\/em>, 1 nolu koltu\u011fa oturmu\u015fsa B <\/em>= 2, C <\/em>= 3, D <\/em>= 4 olacakt\u0131r.<\/p>\n E\u011fer A <\/em>= 2 ise a\u015fa\u011f\u0131daki gibi 4 se\u00e7enek ortaya \u00e7\u0131kar.<\/p>\n A <\/em>= 2, B <\/em>= 1, C <\/em>= 3, D <\/em>= 4,<\/p>\n A <\/em>= 2, B <\/em>= 3, C <\/em>= 1, D <\/em>= 4,<\/p>\n A <\/em>= 2, B <\/em>= 3, C <\/em>= 4, D <\/em>= 1,<\/p>\n A <\/em>= 2, B <\/em>= 4, C <\/em>= 3, D <\/em>= 1.<\/p>\n E\u011fer A <\/em>= 3 ise a\u015fa\u011f\u0131daki gibi 2 se\u00e7enek vard\u0131r.<\/p>\n A <\/em>= 3, B <\/em>= 2, C <\/em>= 1, D <\/em>= 4,<\/p>\n A <\/em>= 3, B <\/em>= 2, C <\/em>= 4, D <\/em>= 1.<\/p>\n E\u011fer A <\/em>= 4 ise B <\/em>= 2, C <\/em>= 3, D <\/em>= 1 olacakt\u0131r.<\/p>\n Dikkat edilirse yukar\u0131da 4 \u00f6\u011frenci i\u00e7in yapt\u0131\u011f\u0131m\u0131z analizde son ki\u015fi olan D<\/em>, hep ya 1 ya da 4 numaral\u0131 koltu\u011fa oturuyor. \u00c7\u00fcnk\u00fc son \u00f6\u011frencinin kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131, salona daha \u00f6nce giren \u00f6\u011frencilerden herhangi birinin 1\u2019inci veya 4\u2019\u00fcnc\u00fc koltu\u011fa oturup oturmad\u0131\u011f\u0131na ba\u011fl\u0131. E\u011fer daha \u00f6nce 1 numaral\u0131 koltu\u011fa oturulmu\u015fsa son ki\u015fi kendi koltu\u011funa, daha \u00f6nce 4 numaral\u0131 koltu\u011fa oturulmu\u015fsa bu kez son \u00f6\u011frenci 1 numaral\u0131 koltu\u011fa oturmak zorunda kal\u0131yor. Bunun nedeniyse 1 veya 4\u2019ten herhangi birine oturuldu\u011funda di\u011fer iki koltu\u011fun (2 ve 3\u2019\u00fcn) dolduruluyor olmas\u0131. Ayr\u0131ca unutmayal\u0131m ki, salona son girecek ki\u015fiden \u00f6nceki \u00f6\u011frencilerin her birinin 1 veya 4 numaral\u0131 koltu\u011fa oturma olas\u0131l\u0131l\u0131\u011f\u0131 ayn\u0131.<\/p>\n \u00c7\u00f6z\u00fcm 1. <\/strong>Yukar\u0131daki \u00e7\u00f6z\u00fcmleme 500 \u00f6\u011frencide de ayn\u0131 olacakt\u0131r. \u00d6rne\u011fin salona ilk giren \u00f6\u011frencinin 200 numaral\u0131 koltu\u011fa oturdu\u011funu varsayal\u0131m. Sonra gelen ilk 198 ki\u015fi (2, 3, 4, 5, \u2026 199) kendi yerlerine oturacaklard\u0131r. E\u011fer 200\u2019\u00fcnc\u00fc ki\u015fi 1 numaral\u0131 koltu\u011fa oturursa 201 ve 201\u2019den sonraki numaral\u0131 \u00f6\u011frenciler kendi yerlerine oturacaklar\u0131ndan 500\u2019\u00fcnc\u00fc \u00f6\u011frenci de kendi koltu\u011funa oturmu\u015f olur. \u015eimdi de 200\u2019\u00fcnc\u00fc ki\u015finin 1 numaraya de\u011fil de 200\u2019den daha b\u00fcy\u00fck numaral\u0131 bir koltu\u011fa oturdu\u011funu varsayal\u0131m. \u00d6rne\u011fin 498 numaral\u0131 koltu\u011fa otursun. Bu durumda sonradan salona giren 297 ki\u015fi (201, 202, 203, \u2026 497) kendi koltuklar\u0131na oturaca\u011f\u0131ndan 498 numaral\u0131 \u00f6\u011frenciye 1, 499 ve 500 numaral\u0131 koltuklar kal\u0131r. B\u00f6ylece<\/p>\n 498 numaral\u0131 \u00f6\u011frenci bu \u00fc\u00e7 koltuktan hangisine oturursa otursun son gelen ki\u015fi ya 1 ya da 500 numaral\u0131 koltu\u011fa oturaca\u011f\u0131ndan istenen olas\u0131l\u0131k \u00bd olur. Buradaki p\u00fcf nokta, yukar\u0131da se\u00e7ti\u011fimiz 498 say\u0131s\u0131n\u0131 daha k\u00fc\u00e7\u00fck alsak da son ki\u015fi salona girmeden \u00f6nce sadece 1 ya da 500 numaral\u0131 koltuklardan sadece birinin bo\u015f kalaca\u011f\u0131d\u0131r.<\/p>\n \u00c7\u00f6z\u00fcm 2. Salona ilk giren \u00f6\u011frenci i\u00e7in 3 farkl\u0131 se\u00e7enek var: ya kendi koltu\u011funa, yani 1 numaral\u0131 yere oturur, ya 500 numaral\u0131 koltu\u011fa ya da ne 1 ne de 500 numaral\u0131 koltuklara oturur. Kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131 1\/500\u2019d\u00fcr. Bu durumda son \u00f6\u011frencinin kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131 1 olur. 500 numaral\u0131 koltu\u011fa<\/p>\n oturursa son \u00f6\u011frencinin kendi yerine oturma olas\u0131l\u0131\u011f\u0131 0 olur. 1 ve 500\u2019den farkl\u0131 bir koltu\u011fa oturma olas\u0131l\u0131\u011f\u0131 498\/500\u2019d\u00fcr. Bu durumda son ki\u015finin kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131 1\/2 olur. Bu se\u00e7eneklerle istenen olas\u0131l\u0131k a\u015fa\u011f\u0131daki gibi hesaplanabilir:<\/p>\n 1\/500.1+1\/500.0+498\/500.1\/2-1\/2<\/p>\n Genel \u00e7\u00f6z\u00fcm 1. Salonda N <\/em>koltuk ve n <\/em>ki\u015fi oldu\u011funu varsayarak herhangi bir ki\u015finin kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131n\u0131 hesaplayal\u0131m. \u00d6nce herhangi bir ki\u015finin kendisine ait olan bir koltu\u011fa oturmama olas\u0131l\u0131\u011f\u0131n\u0131 bulup bu say\u0131y\u0131 1\u2019den \u00e7\u0131karaca\u011f\u0131z.<\/p>\n \u00d6\u011frenciler k\u00fcmesini {A<\/em>1, A<\/em>2, A<\/em>3,\u2026 A<\/em>n<\/em>} ile koltuk numaralar\u0131n\u0131n k\u00fcmesini de {1, 2, 3, \u2026n} ile g\u00f6sterelim.<\/p>\n A<\/em>1 rastgele bir koltu\u011fa oturdu\u011funda, k <\/em>\u2265 2 olmak \u00fczere A<\/em>k <\/em>\u00f6\u011frencisinin kendi numaras\u0131na ait olan koltu\u011fun dolu olma olas\u0131l\u0131\u011f\u0131n\u0131 bulmal\u0131y\u0131z. Daha \u00f6nce de g\u00f6rd\u00fc\u011f\u00fcm\u00fcz gibi A<\/em>k <\/em>\u00f6\u011frencisi kendi koltu\u011funa oturmaya geldi\u011finde 2, 3, \u2026, k<\/em>\u20131 numaral\u0131 koltuklar dolu olacak. Ayr\u0131ca, geriye kalan n<\/em>\u2013k<\/em>+2 koltuktan birine de A<\/em>1<\/em>, A<\/em>2<\/em>, A<\/em>3<\/em>, \u2026, A<\/em>k\u20131 <\/em>\u00f6\u011frencilerinden biri oturacak. O halde k<\/em>\u2019inci koltu\u011fa A<\/em>k <\/em>\u00f6\u011frencisinden ba\u015fka bir \u00f6\u011frencinin oturma olas\u0131l\u0131\u011f\u0131 1\/n-k+2 olur. Buradan A<\/em>k <\/em>\u00f6\u011frencisinin kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131<\/p>\n P(A_(k ) )=1-1\/(n-k+2 olarak bulunur. Yukar\u0131daki kesirde n = k <\/em>al\u0131n\u0131rsa salona son giren \u00f6\u011frencinin kendi koltu\u011funa oturma olas\u0131l\u0131\u011f\u0131 1\/2 olur.<\/p>\n Genel \u00e7\u00f6z\u00fcm 2. <\/strong>Salonda yine N <\/em>koltuk ve n <\/em>ki\u015fi olsun (n <\/em>\u2265 2). E\u011fer, her k <\/em>< n <\/em>i\u00e7in ilk k <\/em>\u00f6\u011frenci kendi koltuklar\u0131na oturursa (grup olarak, tek tek de\u011fil), bundan sonra gelen herkes, son \u00f6\u011frenci de d\u00e2hil, kendi yerine oturur. Ayr\u0131ca, e\u011fer ilk k <\/em>\u00f6\u011frenciden biri son \u00f6\u011frencinin koltu\u011funa oturmu\u015fsa, son \u00f6\u011frenci kendi koltu\u011funa oturamaz. \u015eimdi \u015f\u00f6yle d\u00fc\u015f\u00fcnelim: E\u011fer 1 numaral\u0131 \u00f6\u011frenci numaras\u0131 k<\/em>\u2019den b\u00fcy\u00fck olan bir koltu\u011fa oturmu\u015fsa k<\/em>\u2019inci \u00f6\u011frenci kendi koltu\u011funa oturacakt\u0131r. E\u011fer 1 numaral\u0131 \u00f6\u011frenci numaras\u0131 k<\/em>\u2019den k\u00fc\u00e7\u00fck veya e\u015fit bir koltu\u011fa oturduysa bu durumda k<\/em>\u2019inci \u00f6\u011frencinin se\u00e7enekleri aras\u0131nda 1\u2019inci ve sonuncu koltuklar da olacakt\u0131r. Ki b\u00f6ylece k<\/em>\u2019inci \u00f6\u011frencinin tercihi e\u015fit olas\u0131l\u0131kla yukar\u0131da s\u00f6z\u00fcn\u00fc etti\u011fimiz iki durumdan birine d\u00f6n\u00fc\u015fm\u00fc\u015f olur. Bu \u015fekilde k<\/em>\u2019yi art\u0131rarak N<\/em>\u2019 ye e\u015fitlersek son \u00f6\u011frenci i\u00e7in 1 veya n <\/em>numaral\u0131 koltuklardan ba\u015fka bir se\u00e7ene\u011fin olamayaca\u011f\u0131 sonucuna ula\u015fm\u0131\u015f oluruz.<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" Bu yaz\u0131da k\u0131sa bir hik\u00e2yesi olan ilgin\u00e7 bir olas\u0131l\u0131k sorusunu ele alaca\u011f\u0131z. Bir \u00fcniversitenin mezuniyet t\u00f6reninde \u00fcniversite y\u00f6netimi olas\u0131 protestolar\u0131 engellemek amac\u0131yla her \u00f6\u011frenciye bir numara vererek herkesin kendi numaras\u0131na ait koltu\u011fa oturmas\u0131 kural\u0131n\u0131 koyar. 500 ki\u015filik konferans salonunun giri\u015f kap\u0131s\u0131nda 500 \u00f6\u011frenci numaralar\u0131na g\u00f6re s\u0131ralan\u0131r, ilk s\u0131rada 1, ikinci s\u0131rada 2 numaral\u0131 \u00f6\u011frenci ve […]<\/p>\n","protected":false},"author":375,"featured_media":27365,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[3569,38,514,510],"tags":[208,3587,3588],"class_list":["post-27362","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-173-sayi","category-dergi-sayilari","category-matematik-sohbetleri","category-surekli-bolumler","tag-matematik","tag-mezuniyet","tag-protesto"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/27362","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/users\/375"}],"replies":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/comments?post=27362"}],"version-history":[{"count":0,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/27362\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media\/27365"}],"wp:attachment":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media?parent=27362"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/categories?post=27362"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/tags?post=27362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}