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Hoca, Acil Yayınları'nın fermantasyon, kombinasyon, binom ve olasılık kitabından permütasyon konusunu ele almaktadır.","Videoda permütasyon konusu detaylı olarak anlatılmakta ve çeşitli örnek sorular çözülmektedir. İlk olarak permütasyonun tanımı (n tane elemanın r'li sıralaması) ve faktöriyel kavramı kullanılarak permütasyon formülü (P(n,r) = n! / (n-r)!) açıklanmaktadır. Ardından dört öğrenci yedi koltuğa oturma, çift sayılarla palindron sayı oluşturma, masada kişilerin oturma düzeni ve evli çiftlerin eşler yan yana olmak şartıyla sıralanması gibi çeşitli permütasyon problemleri adım adım çözülmektedir.","Video, AYT sınavına hazırlanan öğrenciler için temel permütasyon sorularının çözüm tekniklerini göstermekte ve günlük hayattan örneklerle konuyu pekiştirmektedir. Hoca, özellikle masanın şekli ve deseni gibi detaylara dikkat edilmesi gerektiğini vurgulamakta ve soru çözüm mantığına odaklanmanın önemini anlatmaktadır."]},"endTime":1439,"title":"Permütasyon Konusu Eğitim Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"Soru Çözümü Serisi Tanıtımı","list":{"type":"unordered","items":["Acil Yayınları'nın fermantasyon, kombinasyon, binom ve olasılık kitabından soru çözümleri yapılacak.","Kitap konu anlatımlı olup, test kısmında zorlanılmayacağı belirtiliyor.","Videolar permütasyon, tekrarlı permütasyon, kombinasyon, binom ve olasılık konularını içerecek."]},"beginTime":11,"endTime":106,"href":"/video/preview/16950019177916090282?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=11&ask_summarization=1"},{"index":1,"title":"Permütasyon Kavramı","list":{"type":"unordered","items":["Permütasyon, n ve r doğal sayıları için n ≥ r olmak üzere n elemanın r'li sıralamasıdır.","n'in r'li permütasyonu P(n,r) = n! / (n-r)! formülüyle hesaplanır.","Permütasyon, bir kümeyi belirli bir sırada düzenlemeyi ifade eder."]},"beginTime":106,"endTime":216,"href":"/video/preview/16950019177916090282?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=106&ask_summarization=1"},{"index":2,"title":"Permütasyon Örnekleri","list":{"type":"unordered","items":["Dört öğrenci yan yana duran yedi koltuğa kaç farklı şekilde oturabilir sorusu çözülerek permütasyon kavramı açıklanıyor.","Aynı konsepte sahip elemanlar varsa, bunları bir bütün olarak düşünebiliriz.","Matematik kitapları bir arada olacak şekilde sıralama örneği veriliyor."]},"beginTime":216,"endTime":552,"href":"/video/preview/16950019177916090282?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=216&ask_summarization=1"},{"index":3,"title":"Soru Çözümüne Başlangıç","list":{"type":"unordered","items":["Palindrom sayılar (soldan sağa ve sağdan sola aynı şekilde okunan sayılar) tanıtılıyor.","Beş basamaklı bir palindrom sayı örneği veriliyor.","A kümesindeki elemanlarla kaç tane çift sayı yazılabilmesi sorusu çözülecek."]},"beginTime":552,"endTime":583,"href":"/video/preview/16950019177916090282?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=552&ask_summarization=1"},{"index":4,"title":"Çift ve Palindrom Sayı Sorusu","list":{"type":"unordered","items":["Çift sayı aranan bir soruda, son basamağa 2 veya 4 yazılabilir, bu nedenle 2 farklı seçenek vardır.","Palindrom sayı olması için ortadaki basamak mecburen 1 olmalıdır.","Geriye kalan basamaklara herhangi bir rakam konulabilir, bu durumda 6 farklı seçenek vardır.","Toplam durum sayısı 6 x 6 = 36 olup, çift sayı olma durumu için 2 farklı seçenek olduğu için toplam 72 durum 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Daha sonra Poisson dağılımına geçilerek, bir mağazaya gelen müşteri sayısı gibi pratik örneklerle konu pekiştirilmektedir.","Video, olasılık konusunu öğrenmek isteyenler için temel formülleri ve çözüm yöntemlerini içermekte, özellikle binom dağılımının sayımla belirtilen kesikli değişkenlerin dağılımı olduğunu ve bir olayın oluş olasılığının büyük ve denek sayısının küçük olduğu durumlarda kullanıldığını vurgulamaktadır."]},"endTime":1081,"title":"Olasılık Dersi: Binom ve Poisson Dağılımları","beginTime":0}],"fullResult":[{"index":0,"title":"Binom Dağılımı Tanıtımı","list":{"type":"unordered","items":["Bu videoda binom ve pozisyon dağılımları ele alınacak, önceki videoda normal dağılım ve z tablosu anlatılmıştı.","Binom dağılımı, sayımla belirtilen kesikli değişkenlerin dağılımıdır ve bir olayın oluş olasılığının büyük, denek sayısının küçük olduğu durumlarda kullanılır.","Binom dağılımında incelenen olayın birbirinden bağımsız iki olası sonucu vardır ve denemeler birbirinden bağımsızdır."]},"beginTime":2,"endTime":59,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=2&ask_summarization=1"},{"index":1,"title":"Binom Dağılım Formülü","list":{"type":"unordered","items":["Binom dağılım formülü n'in r'li kombinasyonu, istenen olayın oluş sayısı ve olma olasılığı ile hesaplanır.","Olma olasılığı (p) ile olmama olasılığı (q) toplamı 1'e eşittir, yani p+q=1.","Formülde olma olasılığı üzeri istenen olayın oluş sayısı ve olmama olasılığı üzeri toplam olay sayısından istenen olayın oluş sayısını çıkararak bulunur."]},"beginTime":59,"endTime":139,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=59&ask_summarization=1"},{"index":2,"title":"Para Atma Örneği","list":{"type":"unordered","items":["Metal ve hilesiz bir para on kez atıldığında, yazı gelme olasılığı 0,5'tir ve gelmeme olasılığı da 0,5'tir.","Bir kez yazı gelme olasılığı 10'un 1'li kombinasyonu çarpı 0,5 üzeri 1 çarpı 0,5 üzeri 9 olarak hesaplanır.","En az iki kez yazı gelme olasılığı, toplamda bir kez veya hiç gelmeme olasılığını çıkararak bulunabilir."]},"beginTime":139,"endTime":404,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=139&ask_summarization=1"},{"index":3,"title":"Çocuk Sayısı Örneği","list":{"type":"unordered","items":["Dört çocuklu bir ailede en az bir erkek olması durumu, 1'den 4'e kadar erkek olma olasılıklarının toplamı olarak hesaplanabilir.","En az bir erkek ve bir kız olması durumu, 1'den 4'e kadar erkek ve kız olma olasılıklarının toplamı olarak hesaplanabilir.","İki yoldan da hesaplanabilir: her bir durumu ayrı ayrı hesaplayıp toplamak veya hiç gelmeme olasılığını çıkarıp toplam ihtimalden bulmak."]},"beginTime":404,"endTime":596,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=404&ask_summarization=1"},{"index":4,"title":"Aile Sayısı Örneği","list":{"type":"unordered","items":["Her biri dört çocuklu 2000 aileden kaçının en az bir erkek çocuğu vardır, bu 15/16 bulunur.","İki erkek olma ihtimali 3/8 olarak hesaplanır.","2000 ailenin içerisinde iki erkek çocuğu olma ihtimali olan aile sayısı 750 olarak bulunur."]},"beginTime":596,"endTime":699,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=596&ask_summarization=1"},{"index":5,"title":"Binom Dağılımı Örnekleri","list":{"type":"unordered","items":["Bir ve iki kız olma ihtimali hesaplanırken, bir kız olma ihtimali 1/2, iki kız olma ihtimali 3/8 olarak bulunur ve toplam 5/8 ihtimal elde edilir.","2000 ailenin bir veya iki kız çocuğu olma ihtimali 1250'dir.","Tüm çocuklarının erkek olma ihtimali 125'tir."]},"beginTime":705,"endTime":809,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=705&ask_summarization=1"},{"index":6,"title":"Pozisyon Dağılımı","list":{"type":"unordered","items":["Pozisyon dağılımı, sayımla belirten değişkenlerin dağılımını ifade eder ve olay sayısı büyük olduğunda olasılık hesaplamak için kullanılır.","Pozisyon dağılımında kullanılan formül: P(X=x) = (e^(-λ)) * (λ^x) / x! şeklindedir.","λ, belirlenen periyotta ortaya çıkan olay sayısıdır ve x ise ortaya çıkma olasılığı araştırılan olaydır."]},"beginTime":809,"endTime":873,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=809&ask_summarization=1"},{"index":7,"title":"Pozisyon Dağılımı Örnekleri","list":{"type":"unordered","items":["Bir mağazaya haftanın belli bir gününde beş dakikada ortalama dört müşteri geldiğinde, beş dakika içerisinde bir müşteri gelme ihtimali 4/e^4 olarak bulunur.","Yarım saatte (otuz dakika) ikiden fazla müşteri gelmesi olasılığı hesaplanırken, beklenen müşteri sayısı altı katına çıkar ve 24 müşteri gelmesi düşünülür.","Yarım saatte ikiden fazla müşteri gelme olasılığı, hiç müşteri gelmeme olasılığı ile bir ve iki müşteri gelme olasılıklarının toplamı olarak hesaplanır."]},"beginTime":873,"endTime":1061,"href":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=873&ask_summarization=1"}],"linkTemplate":"/video/preview/4766786262018287166?parent-reqid=1767134119896845-17187116145036175499-balancer-l7leveler-kubr-yp-sas-254-BAL&text=Solution+Topic&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Binomial and Poisson 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