OLASILIK NED\u0130R?<\/p>\n
<\/p>\n
\u00c7\u0131kt\u0131: Bir deneyde elde edilecek sonu\u00e7lar\u0131n herbirine denir.<\/p>\n
<\/p>\n
Evrensel k\u00fcme: \u00c7\u0131kt\u0131lar\u0131n olu\u015fturdu\u011fu k\u00fcmeye evrensel k\u00fcme denir.Evrensel k\u00fcmeye her eleman 1 kez yaz\u0131l\u0131r. KAHRAMANMARA\u015e kelimesinin harflerini inceleyelim.<\/p>\n
E=(K,A,H,R,M,N,\u015e) s(E)=7<\/p>\n
<\/p>\n
\u00d6rnek uzay: Bir deneyde gelebilecek \u00e7\u0131kt\u0131lar k\u00fcmesine denir.Herbir \u00e7\u0131kt\u0131 ayr\u0131 ayr\u0131 yaz\u0131l\u0131r.<\/p>\n
\u00d6=(K,A,H,R,A,M,A,N,M,A,R,A,\u015e)<\/p>\n
<\/p>\n
Olay: \u00d6rnek uzay\u0131n herbir alt k\u00fcmesine bir olay denir.Yani olmas\u0131 istenen \u00e7\u0131kt\u0131lar\u0131n k\u00fcmesine denir.<\/p>\n
K olma olay\u0131 (K) 1 elemanl\u0131<\/p>\n
A olma olay\u0131 (A,A,A,A,A) 5 elemanl\u0131<\/p>\n
<\/p>\n
Ba\u011f\u0131ml\u0131 olaylar: \u0130ki olaydan herhangi birinin ger\u00e7ekle\u015fmesi di\u011fer olay\u0131n olma olas\u0131l\u0131\u011f\u0131n\u0131 de\u011fi\u015ftiriyorsa bu olaylara ba\u011f\u0131ml\u0131 olaylar denir.<\/p>\n
<\/p>\n
Ba\u011f\u0131ms\u0131z olaylar: \u0130ki olaydan herhangi birinin ger\u00e7ekle\u015fmesi di\u011fer olay\u0131n olma olas\u0131l\u0131\u011f\u0131n\u0131 de\u011fi\u015ftirmiyorsa bu olaylara ba\u011f\u0131ms\u0131z olaylar denir.<\/p>\n
<\/p>\n
Kesin olay: Ger\u00e7ekle\u015fmesi kesin olan olaylara denir. o(A)=1 olan olaylard\u0131r.<\/p>\n
\u00d6rne\u011fin s\u0131nava \u00e7al\u0131\u015fmayan bir \u00f6\u011frencinin s\u0131navdan k\u00f6t\u00fc not almas\u0131 kesin bir olayd\u0131r.<\/p>\n
<\/p>\n
\u0130mkans\u0131z olay: Ger\u00e7ekle\u015fmesi m\u00fcmk\u00fcn olmayan olaylara denir. o(A)=0 olan olaylard\u0131r. \u00d6rne\u011fin bal\u0131\u011f\u0131n kava\u011fa \u00e7\u0131kmas\u0131 imkans\u0131z bir olayd\u0131r.<\/p>\n
<\/p>\n
Olas\u0131l\u0131k: P(A)=S(A) \/ S(E)<\/p>\n
Bir olay\u0131n olas\u0131l\u0131\u011f\u0131=istenilen durumlar\u0131n say\u0131s\u0131 \/ t\u00fcm durumlar\u0131n say\u0131s\u0131<\/p>\n
p(A)=0 ise imkans\u0131z olay=ger\u00e7ekle\u015fmesi m\u00fcmk\u00fcn de\u011fil<\/p>\n
P(A)=1 ise kesin olay=ger\u00e7ekle\u015fmesi kesin<\/p>\n
Herhangi bir olay\u0131n olmama olas\u0131l\u0131\u011f\u0131:<\/p>\n
P'(A) = 1 – P(A)<\/p>\n
<\/p>\n
\u00d6rnek: \u00d6=(M,A,R,M,A,R,A) s(\u00d6)=7<\/p>\n
\u00e7ekilen bir harfin A olma olas\u0131l\u0131\u011f\u0131 O(A)=3\/7<\/p>\n
\u00e7ekilen bir harfin A olmama olas\u0131l\u0131\u011f\u0131 O(A’)=1-3\/7=4\/7<\/p>\n
Ba\u011f\u0131ms\u0131z olay:<\/p>\n
Birbirlerini etkilemiyorlarsa(para-zar)<\/p>\n
P(A \u00c7 B)= P(A) . P(B)<\/p>\n
<\/p>\n
\u00f6rnek: Para ile zar ayn\u0131 anda at\u0131l\u0131yor.Paran\u0131n yaz\u0131, zar\u0131nda 3 gelmesi olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r?<\/p>\n
P(A \u00c7 B)= 1\/2 . 1\/6 = 1\/12<\/p>\n
<\/p>\n
Ayr\u0131k iki olay\u0131n birle\u015fiminin olas\u0131l\u0131\u011f\u0131:<\/p>\n
P(AUB)= P(A) + P(B)<\/p>\n
<\/p>\n
\u00f6rnek: Bir kutuda 1’den 10’a kadar numaraland\u0131r\u0131lm\u0131\u015f 10 kart vard\u0131r.Kutudan rastgele se\u00e7ilen bir kart\u0131n 2 veya 8 numaral\u0131 kart olmas\u0131 olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r?<\/p>\n
P(AUB)= 1\/10 + 1\/10 = 2\/10 = 1\/5<\/p>\n
<\/p>\n
Ayr\u0131k olmayan iki olay\u0131n birle\u015fiminin olas\u0131l\u0131\u011f\u0131:<\/p>\n
P(AUB)= P(A) + P(B) – P(A \u00c7 B)<\/p>\n
<\/p>\n
\u00f6rnek: At\u0131lan bir zar\u0131n \u00fcst y\u00fczeyine gelecek say\u0131lar\u0131n 3’ten b\u00fcy\u00fck veya \u00e7ift gelme olas\u0131l\u0131\u011f\u0131n\u0131 bulunuz?<\/p>\n
E=(1,2,3,4,5,6)<\/p>\n
A=(4,5,6)<\/p>\n
B=(2,4,6)<\/p>\n
A \u00c7 B=(4,6)<\/p>\n
P(AUB)= 3\/6 + 3\/6 – 2\/6 = 4\/6 = 2\/3<\/p>\n
<\/p>\n
Problem: Okan, alfabemizdeki b\u00fct\u00fcn harfleri ayn\u0131 \u00f6zelliklere sahip k\u00e2\u011f\u0131t par\u00e7alar\u0131na yazarak bo\u015f bir kutuya atm\u0131\u015ft\u0131r. Emel, kutudan rasgele bir k\u00e2\u011f\u0131t \u00e7ekmi\u015ftir.<\/p>\n
\u00c7ekilen k\u00e2\u011f\u0131tta \u00fcnl\u00fc harf olma olas\u0131l\u0131\u011f\u0131 nedir?<\/p>\n
Deney: E\u015f \u00f6zelliklere sahip k\u00e2\u011f\u0131t \u00fczerine yaz\u0131lm\u0131\u015f olan alfabemizdeki harflerden birinin se\u00e7ilmesi.<\/p>\n
\u00d6rnek uzay:<\/p>\n
O={alfabemizdeki t\u00fcm harfler} veya<\/p>\n
\u00d6={a,b,c,\u00e7,d,e,f,g,\u011f,h,\u0131,i,j,k,l,m,n,o,\u00f6,p,r,s,\u015f,t,u,\u00fc,v,y,z}, s(\u00d6)=29<\/p>\n
Olay:<\/p>\n
H={bir \u00fcnl\u00fcn\u00fcn \u00e7ekilmesi}veya H={a,e,\u0131,i,o,\u00f6,u,\u00fc},<\/p>\n
s(H)=8<\/p>\n
Olay\u0131n \u00e7\u0131kt\u0131lar\u0131:<\/p>\n
a, e, \u0131, i, o, \u00f6, u, \u00fc<\/p>\n
E\u015f olas\u0131l\u0131kl\u0131 olma: Her bir harfin \u00e7ekilme olas\u0131l\u0131\u011f\u0131 e\u015fittir.<\/p>\n
<\/p>\n
Evrensel k\u00fcmede her bir eleman bir kez yaz\u0131l\u0131r fakat \u00f6rnek uzayda \u00e7\u0131kt\u0131lar ka\u00e7 tane ise o kadar yaz\u0131l\u0131r.<\/p>\n
\u00d6rnek:<\/p>\n
a. \u201cMATEMAT\u0130K\u201d kelimesinin harflerinden olu\u015fan evrensel k\u00fcme: E={M, A, T, E, \u0130, K}<\/p>\n
b. \u201cMatematik\u201d kelimesinin her bir harfi ayn\u0131 \u00f6zelliklere sahip k\u00e2\u011f\u0131t par\u00e7alar\u0131na yaz\u0131larak torbaya at\u0131lm\u0131\u015ft\u0131r.<\/p>\n
\u201cBakmadan bir k\u00e2\u011f\u0131t \u00e7ekildi\u011finde \u00e7\u0131kan harfin \u201cA\u201d olma olas\u0131l\u0131\u011f\u0131 nedir?\u201d sorusundaki \u00f6rnek uzay:<\/p>\n
\u00d6={M, A, T, E, M, A, T, \u0130, K}<\/p>\n
<\/p>\n
<\/p>\n
OLASILIK \u00c7E\u015e\u0130TLER\u0130 NELERD\u0130R?<\/p>\n
<\/p>\n
Deneysel olas\u0131l\u0131k: Bir olas\u0131l\u0131k deneyi sonunda hesaplanan olas\u0131l\u0131\u011fa denir. Bu olas\u0131l\u0131kta deneyin yap\u0131ld\u0131\u011f\u0131 problemin i\u00e7inde ge\u00e7er, problemi okudu\u011funuzda bir \u015feyler yap\u0131ld\u0131\u011f\u0131n\u0131 anlar, verileri g\u00f6r\u00fcrs\u00fcn\u00fcz.<\/p>\n
<\/p>\n
\u00f6rnek: Hileli bir zar 20 kez at\u0131ld\u0131\u011f\u0131nda 3 kez 1, 2 kez 2, 3 kez 3, 2 kez 4, 3 kez 5 ve 7 kez 6 geliyor. Buna g\u00f6re bu zar at\u0131ld\u0131\u011f\u0131nda 5 gelme olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r? cevap: 3\/20<\/p>\n
<\/p>\n
Teorik olas\u0131l\u0131k: Bir olas\u0131l\u0131k deneyinden teorik olarak beklenen olas\u0131l\u0131\u011fa denir.Genelde \u015fimdiye kadar kar\u015f\u0131la\u015ft\u0131\u011f\u0131m\u0131z problem tipleridir.\u0130stenen durumlar\u0131n say\u0131s\u0131n\u0131 tespit edip t\u00fcm durumlara b\u00f6leriz.<\/p>\n
<\/p>\n
\u00f6rnek: Bir zar at\u0131ld\u0131\u011f\u0131nda 3 gelme olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r? cevap: 1\/6<\/p>\n
<\/p>\n
\u00d6znel olas\u0131l\u0131k: Ki\u015filerin kendi d\u00fc\u015f\u00fcncelerine g\u00f6re karar verdikleri olas\u0131l\u0131klara denir.Bu tip problemlerde ki\u015filerin ismi ve tahmini yer al\u0131r.<\/p>\n
<\/p>\n
\u00f6rnek: 25 yumurtadan baz\u0131lar\u0131 \u00e7ift sar\u0131l\u0131d\u0131r.Ali’ye g\u00f6re al\u0131nacak bir yumurtan\u0131n \u00e7ift sar\u0131l\u0131 olma olas\u0131l\u0131\u011f\u0131 10\/25=0,4’t\u00fcr. Ay\u015fe’ye g\u00f6re al\u0131nacak bir yumurtan\u0131n \u00e7ift sar\u0131l\u0131 olma olas\u0131l\u0131\u011f\u0131 15\/25=0,6’d\u0131r.<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
OLASILIK NED\u0130R? \u00c7\u0131kt\u0131: Bir deneyde elde edilecek sonu\u00e7lar\u0131n herbirine denir. Evrensel k\u00fcme: \u00c7\u0131kt\u0131lar\u0131n olu\u015fturdu\u011fu k\u00fcmeye evrensel k\u00fcme denir.Evrensel k\u00fcmeye her eleman 1 kez yaz\u0131l\u0131r. KAHRAMANMARA\u015e kelimesinin harflerini inceleyelim. E=(K,A,H,R,M,N,\u015e) s(E)=7 \u00d6rnek uzay: Bir deneyde gelebilecek \u00e7\u0131kt\u0131lar k\u00fcmesine denir.Herbir \u00e7\u0131kt\u0131 ayr\u0131 ayr\u0131 yaz\u0131l\u0131r. \u00d6=(K,A,H,R,A,M,A,N,M,A,R,A,\u015e) Olay: \u00d6rnek uzay\u0131n herbir alt k\u00fcmesine bir olay … Devam\u0131n\u0131 oku…<\/a><\/p>\n","protected":false},"author":1685,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13],"tags":[14297,14433,14432],"class_list":["post-40357","post","type-post","status-publish","format-standard","hentry","category-mt","tag-olasilik","tag-olasilik-ders-notu","tag-olasilik-nedir"],"_links":{"self":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40357","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/users\/1685"}],"replies":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/comments?post=40357"}],"version-history":[{"count":0,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40357\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/media?parent=40357"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/categories?post=40357"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/tags?post=40357"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}