PERM\u00dcTASYON VE OLASILIK<\/p>\n
<\/p>\n
PERM\u00dcTASYON AMA\u00c7 Perm\u00fctasyonla ilgili temel kavramlar\u0131 kullanabilme becerisi<\/p>\n
Olas\u0131l\u0131k Ama\u00e7 Olas\u0131l\u0131k ve olas\u0131l\u0131kla ilgili temel kavramlar bilgisi<\/p>\n
Planlama Perm\u00fctasyon ve olas\u0131l\u0131k kavram\u0131<\/p>\n
1)Perm\u00fctasyon<\/p>\n
A)Genel \u00e7arpma \u00f6zelli\u011fi<\/p>\n
B) Perm\u00fctasyon<\/p>\n
1) \u201dn\u201d elemanl\u0131 bir k\u00fcmenin n\u2019li perm\u00fctasyonu<\/p>\n
2) \u201cn\u201delemanl\u0131 bir k\u00fcmenin r\u2019li perm\u00fctasyonu<\/p>\n
3)Dairesel perm\u00fctasyon<\/p>\n
2)Olas\u0131l\u0131k:<\/p>\n
A)Olay ve olas\u0131l\u0131k tan\u0131m\u0131<\/p>\n
B)Ayr\u0131k iki olay\u0131n olas\u0131l\u0131\u011f\u0131 (A veya B\u2019nin olas\u0131l\u0131\u011f\u0131)<\/p>\n
C)Ayn\u0131 zamanda ge\u00e7ekle\u015fen ba\u011f\u0131ms\u0131z iki olay\u0131n olas\u0131l\u0131\u011f\u0131(A ve B\u2019nin olas\u0131l\u0131\u011f\u0131)<\/p>\n
\u0130\u015fleni\u015f<\/p>\n
Perm\u00fctasyon ( B\u00fcy\u00fck )<\/p>\n
a) Sayman\u0131n Temel \u0130lkesi ( Genel \u00c7arpma \u00d6zelli\u011fi )<\/p>\n
\u00d6R: Ahmet\u2019in iki de\u011fi\u015fik pantolonu \u00fc\u00e7 de\u011fi\u015fik renk g\u00f6mle\u011fi vard\u0131r.Ahmet g\u00f6mlek ile pantolonunu ka\u00e7 de\u011fi\u015fik bi\u00e7imde giyebilir.<\/p>\n
\u00c7\u00d6Z\u00dcM: Ahmet\u2019in de\u011fi\u015fik renk g\u00f6mlekleri G1,G2,G3 ve pantolonlar\u0131 da P1,P2 olsun.<\/p>\n
Ahmet bu giysileri a\u015fa\u011f\u0131da g\u00f6sterilen bi\u00e7imlerde giyebilir.<\/p>\n
1. Giyinme => G1 P1<\/p>\n
2. Giyinme => G1 P2<\/p>\n
3. Giyinme => G2 P1<\/p>\n
4. Giyinme => G2 P2<\/p>\n
5. Giyinme => G3 P1<\/p>\n
6. Giyinme => G3 P2 bi\u00e7iminde giyebilir.<\/p>\n
Ahmet\u2019in giyini\u015fi 6 de\u011fi\u015fik bi\u00e7imde olmaktad\u0131r. Bunu k\u0131saca,<\/p>\n
<\/p>\n
G\u00f6mlek Pantolon<\/p>\n
3 tane 2 tane<\/p>\n
3 x 2 = 6 \u015feklinde buluruz.<\/p>\n
<\/p>\n
Ard\u0131\u015f\u0131k iki i\u015flemden biri, a de\u011fi\u015fik yoldan yap\u0131labiliyor. Bu yollardan herhangi biri kullan\u0131ld\u0131ktan sonra, ikinci bir i\u015flem b de\u011fi\u015fik yoldan yap\u0131labiliyorsa, ard\u0131\u015f\u0131k iki i\u015flem a x b de\u011fi\u015fik yoldan yap\u0131labilir.<\/p>\n
Bu \u00f6zelli\u011fe \u201cSayman\u0131n Temel \u0130lkesi \u201c yada \u201c Genel \u00c7arpma \u00d6zelli\u011fi\u201d denir.<\/p>\n
<\/p>\n
\u00d6R: A= ( 1,2,3,4,5 } k\u00fcmesinin elemanlar\u0131 ile rakamlar\u0131 farkl\u0131 \u00fc\u00e7 basamakl\u0131 ka\u00e7 \u00e7ift say\u0131 yaz\u0131labilir.<\/p>\n
<\/p>\n
Y O B<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
4 X 3 X 2 = 24 de\u011fi\u015fik \u00e7ift say\u0131 yaz\u0131labilir.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R: A= ( 0,2,3,4,5 } k\u00fcmesinin elemanlar\u0131n\u0131 kullanarak 5 ile b\u00f6l\u00fcnebilen ka\u00e7 tane 3 basamakl\u0131 tek say\u0131 vard\u0131r.<\/p>\n
<\/p>\n
<\/p>\n
Y O B<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
4 X 5 X 1 = 20 tane say\u0131 yaz\u0131labilir.<\/p>\n
<\/p>\n
FAKT\u00d6R\u0130YEL<\/p>\n
<\/p>\n
<\/p>\n
n C N olmak \u00fczere,<\/p>\n
1.2.3. _ _ _ _ _ .n<\/p>\n
\u00e7arp\u0131m\u0131na n fakt\u00f6riyel denir ve<\/p>\n
<\/p>\n
n! = n .(n-1).(n-2)._ _ _ _ _ .3.2.1 bi\u00e7iminde ifade edilir.<\/p>\n
<\/p>\n
0! = 1<\/p>\n
1! = 1<\/p>\n
n! = n.(n-1)! Olarak tan\u0131mlan\u0131r.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
4! = 4.3.2.1 = 24<\/p>\n
5! = 5.4.3.2.1 = 120<\/p>\n
15! 15.14.13!<\/p>\n
13! 13! = 15.14 = 210<\/p>\n
<\/p>\n
4) 8!+9! 8.7!+9.8.7! 7! (8+9.<\/p>\n
7! 7! 7!<\/p>\n
<\/p>\n
8+72 = 80<\/p>\n
<\/p>\n
4!. ( n \u2013 1 )!<\/p>\n
n! = 6 => n = ?<\/p>\n
<\/p>\n
4! . ( n-1 )!<\/p>\n
n! = 6 =><\/p>\n
<\/p>\n
( 4.3.2.1 ) . ( n-1)! = 6 . n!<\/p>\n
<\/p>\n
24. ( n-1)! = 6.n. ( n-1 )!<\/p>\n
<\/p>\n
n 24. ( n-1 )! n = 4<\/p>\n
6 . ( n-1 )!<\/p>\n
<\/p>\n
PERM\u00dcTASYON<\/p>\n
<\/p>\n
<\/p>\n
Bir k\u00fcmenin elemanlar\u0131n\u0131n belli bir s\u0131raya g\u00f6re dizili\u015flerinin her birine bir perm\u00fctasyon denir.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
<\/p>\n
A = ( 1,2,3 } k\u00fcmesinin perm\u00fctasyonlar\u0131n\u0131 yazal\u0131m.<\/p>\n
( 1,2,3 ) ( 2,3,1 )<\/p>\n
( 1,3,2 ) ( 3,1,2 )<\/p>\n
( 2,1,3 ) ( 3,2,1 )<\/p>\n
<\/p>\n
n elemanl\u0131 bir k\u00fcmenin n\u2019li perm\u00fctasyonlar\u0131n\u0131n say\u0131s\u0131 P(n,n) \u015feklinde g\u00f6sterilir.P(n,n) ifadesi, n\u2019den 1\u2019e kadar ard\u0131\u015f\u0131k do\u011fal say\u0131lar\u0131n \u00e7arp\u0131m\u0131d\u0131r.<\/p>\n
<\/p>\n
Yani; P( n,n ) = n! \u2018dir.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R: \u201cAhmet\u201d kelimesinin harfleri ile, 5 harfli anlaml\u0131 yada anlams\u0131z ka\u00e7 kelime yaz\u0131labilir.<\/p>\n
<\/p>\n
P ( 5,5 ) = 5!<\/p>\n
= 5.4.3.2.1<\/p>\n
= 120 bulunur.<\/p>\n
<\/p>\n
<\/p>\n
\u201cn\u201d Elemanl\u0131 Bir k\u00fcmenin \u201cr\u201d li Perm\u00fctasyonlar\u0131<\/p>\n
<\/p>\n
\u201cn\u201d ve \u201cr\u201d birer sayma say\u0131s\u0131 ( n > r ) olmak \u00fczere , n elemanl\u0131 bir k\u00fcmenin elemanlar\u0131n\u0131n r\u2019li s\u0131ralan\u0131\u015f\u0131na, \u201c n elemanl\u0131 k\u00fcmenin r\u2019li perm\u00fctasyonu \u201c denir.ve<\/p>\n
P ( n,r ) \u015feklinde g\u00f6sterilir.<\/p>\n
P ( n,r ) perm\u00fctasyonlar\u0131n\u0131n say\u0131s\u0131,<\/p>\n
<\/p>\n
<\/p>\n
P ( n,r ) = n! \u0130fadesi ile bulunur.<\/p>\n
( n-r )!<\/p>\n
<\/p>\n
Ba\u015fka bir ifadeyle P ( n,r ) perm\u00fctasyonlar\u0131n\u0131n say\u0131s\u0131n\u0131 bulmak i\u00e7in, n\u2019den geriye do\u011fru, r tane ard\u0131\u015f\u0131k \u00e7arpan \u00e7arp\u0131l\u0131r.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
1) P ( 5,2 ) 5! 5.4.3! = 20<\/p>\n
( 5-2 )! 3!<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
2) P ( 7,3 ) 7! 7.6.5.4! = 210<\/p>\n
( 7-3 )! 4!<\/p>\n
<\/p>\n
<\/p>\n
3) P ( 6,1 ) 6! 6.5! = 6<\/p>\n
( 6-1 )! 5!<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
P ( 5,3 ) = 5.4.3 = 60<\/p>\n
P ( 6,2 ) = 6.5.4.3.2 = 720<\/p>\n
P ( 7,4 ) = 7.6.5.4 = 840<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R: 5. P( n,3 ) = 2. P( n+1,3 ) e\u015fitli\u011finde n\u2019nin de\u011feri ka\u00e7t\u0131r?<\/p>\n
<\/p>\n
\u00c7\u00d6Z\u00dcM:<\/p>\n
<\/p>\n
<\/p>\n
5.n ( n-1 ). ( n-2 ) = 2.( n+1 ) n . ( n-1 )<\/p>\n
<\/p>\n
5.( n-2 ) = 2 ( n+1 )<\/p>\n
<\/p>\n
5 n-10 = 2 n+2<\/p>\n
<\/p>\n
5 n \u20132n = 2+10<\/p>\n
<\/p>\n
3 n = 12<\/p>\n
<\/p>\n
n = 4<\/p>\n
<\/p>\n
D\u00f6nel (Dairesel ) S\u0131ralama<\/p>\n
<\/p>\n
\u201cn\u201d elemanl\u0131 bir k\u00fcmenin elemanlar\u0131n\u0131n, bir \u00e7emberin noktalar\u0131 \u00fczerinde birbirine g\u00f6re farkl\u0131 dizili\u015flerinden her birine,\u201ddairesel perm\u00fctasyon \u201c denir.<\/p>\n
\u201cn\u201d elemanl\u0131 bir k\u00fcmenin elemanlar\u0131n\u0131n, bir daire \u00fczerinde de\u011fi\u015fik bi\u00e7imde dairesel perm\u00fc-<\/p>\n
tasyonlar\u0131n\u0131n say\u0131s\u0131,<\/p>\n
<\/p>\n
( n-1 )! Tanedir.<\/p>\n
<\/p>\n
\u00d6R: 7 ki\u015fi, yuvarlak bir masan\u0131n etraf\u0131nda ka\u00e7 de\u011fi\u015fik \u015fekilde oturabilir?<\/p>\n
<\/p>\n
\u00c7\u00d6Z\u00dcM:<\/p>\n
Bir ki\u015finin yeri sabit tutulursa;<\/p>\n
<\/p>\n
Oturu\u015f say\u0131s\u0131 = ( 7-1 )!<\/p>\n
= 6!<\/p>\n
6.5.4.3.2.1 = 720 bulunur.<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
Bir okulda, 3 y\u00f6netici ile 5 \u00f6\u011fretmen vard\u0131r. Y\u00f6neticiler yan yana olmak \u00fczere, 8 ki\u015fi yuvarlak bir masan\u0131n etraf\u0131na oturacaklard\u0131r. Oturu\u015f bi\u00e7imi ka\u00e7 farkl\u0131 bi\u00e7imde olabilir?<\/p>\n
<\/p>\n
\u00c7\u00d6Z\u00dcM:<\/p>\n
<\/p>\n
Y\u00f6neticiler bir arada olaca\u011f\u0131 i\u00e7in, \u00fc\u00e7 y\u00f6neticiyi bir ki\u015fi gibi kabul edelim.<\/p>\n
Bu duruma g\u00f6re, yuvarlak masan\u0131n etraf\u0131na 1+5 = 6 ki\u015fi oturuyormu\u015f gibi d\u00fc\u015f\u00fcnebiliriz. Ancak,3 y\u00f6netici de kendi aralar\u0131nda 3! Kadar farkl\u0131 bi\u00e7imde otururlar.<\/p>\n
<\/p>\n
Buna g\u00f6re, farkl\u0131 oturu\u015f bi\u00e7imi,<\/p>\n
<\/p>\n
3!.( 6-1 )! = 6 .120 =720 de\u011fi\u015fik bi\u00e7imde olur.<\/p>\n
<\/p>\n
<\/p>\n
OLASILIK<\/p>\n
<\/p>\n
<\/p>\n
Olas\u0131l\u0131k, rastlant\u0131 yada kesin olmayan olaylarla u\u011fra\u015f\u0131r. Rastlant\u0131; sonucu \u00f6nceden bilinmeyen, ger\u00e7ekle\u015fmesi \u015fansa ba\u011fl\u0131 olaylard\u0131r.<\/p>\n
\u00d6rne\u011fin; bir paray\u0131 havaya att\u0131\u011f\u0131m\u0131zda, yaz\u0131 m\u0131 yoksa tura m\u0131 gelece\u011fini deney yapmadan bilemeyiz.<\/p>\n
<\/p>\n
. Bir deneyde \u00e7\u0131kan sonu\u00e7lar\u0131n her birine \u201c olay \u201cdenir. Yap\u0131lan bir deneyde, elde edile-<\/p>\n
bilecek t\u00fcm \u00e7\u0131kanlar\u0131n k\u00fcmesine \u201c\u00f6rnek uzay\u201dveya \u201c evrensel k\u00fcme \u201c ad\u0131 verilir. B\u00fcy\u00fck \u201cE\u201dharfi ile g\u00f6sterilir.<\/p>\n
<\/p>\n
. Bir olay her zaman olabiliyorsa buna \u201ckesin olay\u201d; hi\u00e7 ger\u00e7ekle\u015fmiyorsa buna da \u201cimkans\u0131z olay\u201d denir.<\/p>\n
<\/p>\n
. Bir E \u00f6rnek uzay\u0131n\u0131n her eleman\u0131n\u0131n elde edilme olas\u0131l\u0131\u011f\u0131 e\u015fit ise bu E \u00f6rnek uzay\u0131na \u201ce\u015f olumlu \u00f6rnek uzay \u201c denir.<\/p>\n
<\/p>\n
E\u015f olumlu \u00f6rnek uzay\u0131na ait bir A olay\u0131n\u0131n olas\u0131l\u0131\u011f\u0131 P( A ) bi\u00e7imde g\u00f6sterilir.<\/p>\n
<\/p>\n
A C E olay\u0131 i\u00e7in,<\/p>\n
<\/p>\n
P( A ) = s( A)<\/p>\n
s( E ) dir.<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
Bir zar at\u0131ld\u0131\u011f\u0131nda,\u00fcste gelen y\u00fcz\u00fcn\u00fcn asal say\u0131 olma olas\u0131l\u0131\u011f\u0131 nedir?<\/p>\n
\u00c7\u00d6Z\u00dcM:<\/p>\n
Evrensel k\u00fcme E = ( 1,2,3,4,5,6 }<\/p>\n
Olay A = ( 2,3,5 } dir.<\/p>\n
<\/p>\n
A olay\u0131n\u0131n olas\u0131l\u0131\u011f\u0131 : P( A ) s( A ) 3 1<\/p>\n
s( E ) 6 2<\/p>\n
\u00d6ZELL\u0130KLER<\/p>\n
<\/p>\n
<\/p>\n
Bir olay\u0131n olas\u0131l\u0131\u011f\u0131, s\u0131f\u0131r ile bir aras\u0131nda bir say\u0131d\u0131r.<\/p>\n
<\/p>\n
0 < P( A ) < 1<\/p>\n
P( A ) = 0 => b\u00f6yle bir olaydan s\u00f6z edilemez.(\u0130mkans\u0131z olay )<\/p>\n
P( A ) = 1 => olas\u0131l\u0131k tamd\u0131r.( Kesin olay )<\/p>\n
Bir olay\u0131n olma olas\u0131l\u0131\u011f\u0131 ile olmama olas\u0131l\u0131\u011f\u0131n\u0131n toplam\u0131 1\u2019e e\u015fittir.<\/p>\n
P( A ) + P( A\u2018) = 1 dir.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
Bir torbada,ayn\u0131 b\u00fcy\u00fckl\u00fckte 3 k\u0131rm\u0131z\u0131, 4 beyaz, 5 mavi bilye vard\u0131r. Torbadan rasgele bir bilye \u00e7ekiliyor. \u00c7ekilen bilyenin beyaz olma olas\u0131l\u0131\u011f\u0131 nedir?<\/p>\n
<\/p>\n
\u00c7\u00d6Z\u00dcM:<\/p>\n
<\/p>\n
\u00d6rnek uzay\u0131n eleman say\u0131s\u0131,<\/p>\n
<\/p>\n
s( E ) = 3+4+5 = 12 dir.<\/p>\n
<\/p>\n
Beyaz bilye \u00e7ekme olay\u0131 B olsun . Torbada 4 tane beyaz bilye oldu\u011fundan,<\/p>\n
s( B ) = 4 t\u00fcr. Buna g\u00f6re;<\/p>\n
P( B ) s( B ) 4 1<\/p>\n
s( E ) 12 3 t\u00fcr.<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
Bir \u00e7ift zar, ayn\u0131 anda masan\u0131n \u00fczerine at\u0131l\u0131yor. \u00dcste gelen say\u0131lar\u0131n toplam\u0131n\u0131n asal say\u0131 olma olas\u0131l\u0131\u011f\u0131 nedir?<\/p>\n
<\/p>\n
\u00c7\u00d6Z\u00dcM:<\/p>\n
<\/p>\n
Evrensel k\u00fcmenin eleman say\u0131s\u0131,<\/p>\n
<\/p>\n
s( E ) = 6 x 6 = 36 d\u0131r.<\/p>\n
<\/p>\n
\u00dcste gelen say\u0131lar\u0131n toplam\u0131n\u0131n asal say\u0131 olma durumlar\u0131;<\/p>\n
A = ( (1,1),( 1,2 ),( 2,1 ),( 1,4 ),( 4,1 ),( 1,6 ),( 6,1 ),( 2,3 ),( 3,2 ),( 2,5 ),( 5,2 ),( 3,4 ),( 4,3 ), ( 5,6 ),( 6,5 )}<\/p>\n
<\/p>\n
<\/p>\n
P( A ) s( A) 15 5<\/p>\n
s(E) 36 12 dir.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
AYRIK \u0130K\u0130 OLAYIN B\u0130RLE\u015eMELER\u0130N\u0130N ( A VEYA B OLAYININ ) OLASILI\u011eI<\/p>\n
<\/p>\n
<\/p>\n
Ayr\u0131k olaylar\u0131n birle\u015fimlerinin olas\u0131l\u0131\u011f\u0131, bu olaylar\u0131n olas\u0131l\u0131klar\u0131 toplam\u0131na e\u015fittir.<\/p>\n
A n B = O =><\/p>\n
P ( A U B ) = P ( A ) + P( B ) dir.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
Bir torbaya ayn\u0131 b\u00fcy\u00fckl\u00fckte 2 k\u0131rm\u0131z\u0131, 3 sar\u0131,4 mavi bilye konuluyor. Torbadan rasgele bir bilye \u00e7ekilirse,\u00e7\u0131kan bilyenin k\u0131rm\u0131z\u0131 veya mavi olma olas\u0131l\u0131\u011f\u0131 nedir?<\/p>\n
<\/p>\n
\u00c7\u00d6Z\u00dcM: Evrensel k\u00fcme,<\/p>\n
<\/p>\n
E = ( k1,k2,s1,s2,s3,m1,m2,m3,m4 }ve s( E ) = 9 dur.<\/p>\n
K\u0131rm\u0131z\u0131 bilyeler = A = ( k1,k2 }<\/p>\n
Mavi bilyeler = B = ( m1,m2,m3,m4 }<\/p>\n
A n B = O dir. Buna g\u00f6re,<\/p>\n
P ( A U B ) = P( A ) + P( B ) yaz\u0131l\u0131r.<\/p>\n
P ( A U B ) = 2 4 6 2<\/p>\n
9 9 9 3 bulunur.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
AYRIK OLMAYAN \u0130K\u0130 OLAYIN B\u0130RLE\u015e\u0130MLER\u0130N\u0130N (A VEYA B OLAYININ )<\/p>\n
OLASILI\u011eI<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
Ayr\u0131k olmayan iki olay\u0131n birle\u015fimlerinin olas\u0131l\u0131\u011f\u0131, bu olaylar\u0131n ayr\u0131 ayr\u0131 olas\u0131l\u0131klar\u0131 toplam\u0131ndan kesi\u015fimlerinin olas\u0131l\u0131\u011f\u0131n\u0131n fark\u0131na e\u015fittir.<\/p>\n
<\/p>\n
A n B = O => ,<\/p>\n
<\/p>\n
P ( A U B ) = P( A ) + P( B ) \u2013 P( A n B ) dir.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
\u00d6R:<\/p>\n
<\/p>\n
Bir torbaya 1\u2019den 9\u2019a kadar numaralanm\u0131\u015f ayn\u0131 b\u00fcy\u00fckl\u00fck ve \u00f6zellikte 9 top konuyor. Torbadan rasgele bir top \u00e7ekiliyor. 4\u2019ten b\u00fcy\u00fck veya tek numaral\u0131 bir topun \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131 nedir?<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
\u00c7\u00d6Z\u00dcM:Evrensel k\u00fcme<\/p>\n
E = ( 1,2,3,4,5,6,7,8,9 }<\/p>\n
s( E ) = 9 dur.<\/p>\n
<\/p>\n
Tek numaral\u0131 bilyenin \u00e7\u0131kmas\u0131 olay\u0131;<\/p>\n
A = ( 1,3,5,7,9 }, s( A ) = 5 \u2019tir.<\/p>\n
<\/p>\n
4 ten b\u00fcy\u00fck numaral\u0131 bilyenin \u00e7\u0131kmas\u0131 olay\u0131;<\/p>\n
<\/p>\n
B = ( 5,6,7,8,9 }, s ( B ) = 5\u2019tir.<\/p>\n
A n B = (5,7,9 }, s ( A n B ) = 3\u2019t\u00fcr.<\/p>\n
<\/p>\n
<\/p>\n
P( A ) = 5 P ( B ) = 5 P( A n B) = 3<\/p>\n
9 9 9 dur.<\/p>\n
<\/p>\n
Buna g\u00f6re,<\/p>\n
<\/p>\n
<\/p>\n
P( A u B ) = P( A ) + P( B )- P( A n B )<\/p>\n
= 5 + 5 – 3<\/p>\n
9 9 9<\/p>\n
= 7<\/p>\n
olur.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
BA\u011eIMSIZ OLAYLARIN B\u0130RL\u0130KTE OLMA ( A VE B OLAYININ ) OLASILI\u011eI<\/p>\n
<\/p>\n
<\/p>\n
\u0130ki veya daha \u00e7ok olay\u0131n ger\u00e7ekle\u015fmeleri birbirine ba\u011fl\u0131 de\u011filse b\u00f6yle olaylara \u201c ba\u011f\u0131ms\u0131z olaylar\u201d denir.<\/p>\n
Ba\u011f\u0131ms\u0131z olaylar\u0131n birlikte olma olas\u0131l\u0131\u011f\u0131 bu olaylar\u0131n olas\u0131l\u0131klar\u0131n\u0131n \u00e7arp\u0131m\u0131na e\u015fittir.<\/p>\n
<\/p>\n
<\/p>\n
P( A ve B ) = P( A n B ) = P( A ) . P( B ) dir.<\/p>\n
<\/p>\n
\u00d6R: Bir okulun birinci s\u0131n\u0131f\u0131nda 12 erkek ve 8 k\u0131z, ikinci s\u0131n\u0131f\u0131nda 6 erkek ve 12 k\u0131z \u00f6\u011frenci vard\u0131r. Her iki s\u0131n\u0131ftan da rasgele se\u00e7ilen birer \u00f6\u011frencinin ikisinin de k\u0131z \u00f6\u011frenci olma olas\u0131l\u0131\u011f\u0131 nedir?<\/p>\n
s\u0131n\u0131ftan se\u00e7ilen \u00f6\u011frencinin k\u0131z \u00f6\u011frenci olmas\u0131 olay\u0131 A =><\/p>\n
P( A) = s( A ) 8 2<\/p>\n
s( E ) 20 5 tir.<\/p>\n
s\u0131n\u0131ftan se\u00e7ilen \u00f6\u011frencinin k\u0131z \u00f6\u011frenci olmas\u0131 olay\u0131 B =><\/p>\n
P( B ) = s( B) 12 2<\/p>\n
s( E ) 18 3 t\u00fcr.<\/p>\n
P( A n B ) = P( A ) . P( B )<\/p>\n
= 2 . 2 4<\/p>\n
5 3 15 olur…<\/p>\n","protected":false},"excerpt":{"rendered":"
PERM\u00dcTASYON VE OLASILIK PERM\u00dcTASYON AMA\u00c7 Perm\u00fctasyonla ilgili temel kavramlar\u0131 kullanabilme becerisi Olas\u0131l\u0131k Ama\u00e7 Olas\u0131l\u0131k ve olas\u0131l\u0131kla ilgili temel kavramlar bilgisi Planlama Perm\u00fctasyon ve olas\u0131l\u0131k kavram\u0131 1)Perm\u00fctasyon A)Genel \u00e7arpma \u00f6zelli\u011fi B) Perm\u00fctasyon 1) \u201dn\u201d elemanl\u0131 bir k\u00fcmenin n\u2019li perm\u00fctasyonu 2) \u201cn\u201delemanl\u0131 bir k\u00fcmenin r\u2019li perm\u00fctasyonu 3)Dairesel perm\u00fctasyon 2)Olas\u0131l\u0131k: A)Olay ve olas\u0131l\u0131k tan\u0131m\u0131 B)Ayr\u0131k iki olay\u0131n … 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":[],"categories":[13],"tags":[14297,14296],"_links":{"self":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40167"}],"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=40167"}],"version-history":[{"count":0,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/40167\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/media?parent=40167"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/categories?post=40167"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/tags?post=40167"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}