CHUONG 2: BitM c6 - XAC XUAT CUA Bif M c6
III. CAC BAI TOAN LUYEN TAP
^80 ^72 563 2054
So cham So Ian xuat h i ^ n
1 14
2 18
3 30
4 12
5 14
12
Cty TNHH MTV DVVH Khang Vii^t
Hay t i m xac suat ciia cac bien co: , A: "mat sau cham xuat hi?n" , ; . ,,, ,
B: " mat hai cham xuat h i ^ n "
C: " mQt m|it le xualt hi?n". (':f}n < , .: i -{/-/fl Giai
Xem vi?c tung con siic sic la mot phep t h u ngau nhien. /•
So Ian thyc hi^n phep thu: N = 100
12 3 1
So'Ian xuat hien cua bien co A: 12. Suy ra : P( A) = So Ian xuat hien cua bien CO B: 18. Suy ra P(B)
100 25 18 9 . 100 50
So Ian xuat hi^n ciia bien co C: 14 + 30 +14 = 58. Suy ra P(C) = 58 29 100 50 Bai 2. Tung mQt dong tien hai Ian. T i m xac suat de hai Ian tung do:
1. Deu la mat S 2. M o t S mot N . ^ ^
Giai ''i^-^
Ta CO khong gian mau Q = {SS,SN,NN,NS} => n(Q) = 4 " MP^' AmT r Gpi cac bien co: A: " hai Ian tung deu la mat sap"
B: " hai Ian tung co mot S mot N ' Suy ra A = {SS} ^ n( A) = 1; B = {SN, NS| =^ n(B) = 2
n(A) ^ 1 . n(Q) 4 ' ' n(B) _ 2 _ 1
n(Q) ~ 4 " 2
Bai 3. M p t binh d u n g 16 vien bi, 7 vien bi trang, 6 vien bi den, 3 vien bi do.
1. Lay ngau nhien ba vien bi. Tinh xac suat cua cac bien co': . A: "Lay duoc 3 vien d o " ,, j >'J'^)n. d±i.
B: " Lay ca ba vien bi khong co bi d o "
C: " Lay dugc 1 bi trang,! bi den,l bi do".
2. Lay ngau nhien 4 vien bi.Tinh xac suat cua cac bien co :
X: "Lay diing 1 vien bi trang" ••
Y: " Lay dung 2 vien bi trSng"
3. Lay ngau nhien 10 vien bi. Tinh xac suat ciia bien co D: "lay duoc 5 vien bi trang, 3 bi den, 2 bi d o " .
^ 151
Phdn /(XII pillion^ jtlia/i^•^lui Dai so—Giai tich It Giai 1. Taco; n(Q) = C^^ =560
n ( A ) = C ^ = l ^ P ( A ) = - i - ; n(B) = C^j = 286 P(B) = ^
DoO 280
^' n(C) = C^C^C^ = 1 2 6 ^ P ( C ) = - , . A , ...
2. T a c 6 : n ( n ) = C^^=1820
n(X) = C\ = 588 ^ P(X) = ^ ; n(Y) = C^.C^ = 756 ^ P(Y) = — .
65 65 3. Ta c6: n(Q) = C j " = 8008 \
n{D) = C ^ C ^ C | = 1 2 6 0 ^ P ( D ) = ; ^ . ^ •
Bai 4. Trong m p t chie'c hpp c6 7 vien bi trang, 8 vien b i do va 10 vien b i vang.
Lay ngau nhien ra 6 vien bi
1. T i n h so phan t u ciia khong gian mau. f?j
2. T i n h xac sua't cua cac bien CO sau: ,, i , . i ô , , A: " 6 vien b i lay ra cung mot m a u " ! ; |
B: " CO i t nhat m o t vien bi mau vang" \ ffx ^:,.A •m •^(mi C: " 6 vien bi lay ra c6 d i i ba m a u " .
. , „ > , , , • Giai „ V ;
447 1. Ta c6: n(n) = = 177100 • „ 1 2. Taco: n ( A ) = C ^ + c f = 2 4 5 = ^ P ( A ) = —
'" 5060 Ta c6: n(B) = C j j ^ n(B) = C^g - C^j = 172095 => P(B) =
Ta c6: So'each lay 6 vien bi cung mpt mau: 245 each.
So each lay 6 vien bi g o m hai mau:
C^s - ( C ^ ) + C% -(Cto + ) + -(C^o ^Cl) = 35455 .
Suy ra n(C) = 177100 - 35455 - 245 = 141400. Vgy P(C) = 202 ' 253" / f t . l X 152
Cty r N H H M T V D VVH Kluutx Vift B a i S . C h p n ngau nhien 5 quan bai trong c6 bai tu lo kha. Tinh xac sua't de
trong sap bai chua hai bp doi (hai con cung thupc 1 bp, hai con thupc bo t h u 2, con thu 5 thupc bp khae).
Giai
Gpi A la bien c6'each chpn thoa man yeu eau bai toan.
Chpn hai bp 2 co Cj3 each, m o i bp c6 C4 each vay c6 :
Cj3.C^.C^ each . CO n each chpn bp 1 . :
M o i each chpn bp 1 co 4 each chpn vay c6 Cj3.C4.C4.n.4 = n(A);
n(Q) = C^2 • Vay P(A) = 198 11,/;
4165 'iff i'lip.^r'.ci'''.^ •"'•'>' i:^y'-''^-,)-'ry< ' -
Bai 6. M p t doan tau co 7 toa 6 m p t san ga. Co 7 hanh khach t u san ga len tau, moi nguoi doe lap v o i nhau va chpn mpt toa mpt each ngau nhien. T i m xac suat ciia cac bien co sau:
A: " M p t toa 1 nguoi, m p t toa 2 nguoi, mpt toa co 4 nguoi len va bon toa khong CO n g u o i nao ca" 1
B: " M o i toa CO d i i n g mpt nguoi len". • • Giai
So each len toa cua 7 nguoi la: Q =7^ . 1. T i n h P(A) = ? n ' > i i •
Ta t i m so kha nang thuan Ipi cua A n h u sau:
• Chpn 3 toa CO nguoi len: A7
• V o i toa CO 4 nguoi len ta co: C7 each chpn
• V o i toa CO 2 nguoi len ta co: C3 each chpn
• N g u o i cuoi cung cho vao toa eon lai nen co 1 each
:v,: •[,., >...ijig'n' (hhl ô<>t.r:
Theo quy tac nhan ta co: Q^j = A7.C7.C3 . Do do: P(A) = 450 Q 16807 2. T i n h P(B) = ? ' • M- 1 ' • • ' r^-
M o i m p t each len toa thoa man yeu eau bai toan chinh la m p t hoan v j cua 7 phan t u nen ta eo: fig = 7!
Do do: P(B) =
Q 7' T^ryni' 77
Bai 7. M p t nguoi bo ngau nhien bon la t h u vao 4 bi t h u da dupe ghi dja chi. Tinh xac sua't ciia cac bien co A: " Co it nha't mpt la t h u bo dung phong bi cua no".
Pliiht loiii va phtfmtg phiiptgiiii Dai no'- Gii'ii tick 11 Giai
So each bo 4 la thu- vao 4 bi thir la: |Q| = 4 ! = 24
K i hieu 4 la t h u la: L j , L 2, L 3, L 4 va bo (L , , L2 ,L3 ,L 4 ) la mot hoa v i cua cac so 1,2,3,4 trong do L, = i (i = 1,4 ) neu la t h u bo d u n g dja chi. ^ , Ta xet cac kha nang sail; , r .i-
• Co 4 la t h u bo d u n g dia chi: (1,2,3,4) nen c6 1 each bo
• Co 2 la t h u bo d u n g dja chi: * ' ''^^^ t ' >
;(A)> + ) S o c a c h b 6 2 l a t h u d u n g d i a c h i l a : C^ i ot>rt^ ,'>:',' I' i H +) K h i do CO 1 each bo hai la t h u con lai
Nen t r u a n g hop nay c6: C^ = 6 each bo.
Co diing 1 la t h u bo d u n g dja chi:
r So each chon la t h u bo d u n g dia chi: 4 each So'each chon bo ba la t h u con lai: 2.1 = 2 each Nen t r u o n g hop nay c6: 4.2 = 8 each bo.
D o d o : n. : 1 + 6 + 8 = 15 . Vay P ( A ) = = — = -
^ ' Q 24 8
Bai 8. Gieo mot con xiic sic dong chat can doi ba Ian lien tie'p. T i m xac sua't cua cac bien co sau:
A : " Tong so cham xua't hien trong ba Ian la 1 0 " V - ( A j U f( i 8ô l B: "Co it nhat mot mat chan xua't hien". f!4 o;rrrtiit5T
Giai Ta CO cac kha nang xay ra la: Q = 6.6.6 = 216
1. Goi day ( x ^ X j , X 3) la ke't qua theo t h u t u cua ba Ian gieo voi
^ X j , X 2 , X 3 G { 1 , 2 , 3 , 4 , 5 , 6 } . '>'"^>'^'>ô'" 5''>ằ*^
Phuong trinh x^ + X2 + X3 = 1 0 co cac bo nghiem (chua tinh hoan vj) la:
( 1 , 3 , 6 ) ; ( 1 , 4 , 5 ) ; ( 2 , 2 , 6 ) ; ( 2 , 3 , 5 ) ; ( 2 , 4 , 4 ) ; ( 3 , 4 , 3 ) .
Vai moi bo nghiem ba so phan biet cho ta 3 ! = 6 kha nang xay ra, con cac bo
nghiem (2,2,6); ( 2 , 4 , 4 ) va ( 3 , 4 , 3 ) chi co ba kha nang xay ra S D o d o = 6.3 + 3.3 = 27 nen P ( A ) =
2. Kha nang xua't hien mat le ciia moi Ian gieo la: 3
Suy ra kha nang ba Ian gieo deu xua't hien mat le la: 3"^ = 27 D o d o Q B = 2 1 6 - 2 7 = 189 nen P ( B ) = Op 189 7
2 1 6 ~ 8
•m
C t y TNHH MTV DVVH Khang Vipt
Van dl 3.6ae quy tat tinh xac suat
A P H U O N G P H A P 1 , / - 1. Q u y tac cong xac suat -
Neu hai bien co A va B xung khSc thi P( A u B) = P( A ) + P(B)
I • M a rong quy tac cong xac sua't i . i „ ' Cho k bien CO A ] , A2,...,A|^ doi mot xung khac. K h i do: ulv :
P( A , u A2 u . . . u A J = P( A i ) + P( A 2 ) + ... + P( A , ) . ,^ • ; jM-i , ,
• P( A ) = 1 - P( A ) _ , •' • :
• Giai su A va B la hai bien co tiiy y cung lien quan den mot phep thu.
L i i c d o : P( A u B ) = P( A ) + P( B) - P( A B ) .
2. Q u y tac nhan xac suat ,ô ici . fTwt
^||||, • Ta noi hai bien co A va B doc lap neu su xay ra (hay khong xay ra) cua A khong lam anh huong den xac sua't cua B.
• Hai bien co A va B doc lap khi va chi khi P ( A B ) = P ( A ) . P ( B )
B. B A I T O A N 1 : T I N H X A C S U A T B A N G Q U Y T A C C O N G ,d i o O £ I. P H U O N G P H A P : Su d u n g cac quy tac de'm va cong thuc bien co' do'i, cong
thue bien co hop.
• P (A u B) = P (A ) + P(B) voi A va B la hai bien co xung khac ''T
• P ( A ) = 1 - P ( A ) . , ,,j%mxu?t
I I . C A C v i D V
V i du 1. M o t con sue sac khong dong chat sao cho mat bon cham xua't hien nhieu gap 3 Ian mat khac, cac mat eon lai dong kha nang. T i m xac sua't de xua't hien mot mat c h i n . • I ] [ -'j ' ;^ ô^''^^'^
Giai : Goi Aj la bien C O xua't hien mat i cha'm (1 = 1,2,3,4,5,6)
Ta CO P( A i ) = P( A 2 ) = P ( A 3 ) = P( A 5 ) = P(Af,) = - ! - P ( A 4 ) = x ^(^^" ^ Do ^ P6 1 ( A k ) = 1 ^ 5 x + 3x = l : r ^ x = - ! i ' ;
Goi A la bien CO xua't hien mat chan, suy ra A = A2 U A 4 u A ^ .
- - 1 3 1 5 V I cac bien CO A^ xung khac nen: P(A) = P(A2) + P(A4) + P(A6) = - + - + - = - •
Phan loai P<? phuamg phap giai Bat so-Giai tick 11
V i d u 2. Gieo m o t con xuc sac 4 Ian. T i m xac suat ciia bie'n,c6':
A: " M a t 4 cham xua't hien it nha't mot I a n "
! B: " M a t 3 cham xuat hien d u n g mot Ian"
G i a i
1, Goi Aj la bie'n co " mat 4 cham xua't hien Ian t h u i " v o i i = 1,2,3,4 . K h i do: Aj la bie'n co " M a t 4 cham khong xua't hien Ian t h u i " j Va p(A,) = l - P ( A i ) = l - - = ^ 1 = ^ •. -.N''
6 6
Ta co: A la bie'n co: " khong co mat 4 cham xua't hien trong 4 Ian gieo"
Va A = A j . A 2. A 3 . A 4 . V i cac Aj dpc lap voi nhau nen ta co:
6 j
; ; P ( A ) = P(A I)P(A 2 ) P { A 3 ) P(A 4 ) =
• Vay P ( A ) = 1 - P(A) = 1 - ^ - J . .,,;!-u^.,,.0^^^^^^^^^^ ' .
2. Goi Bj la bie'n co " mat 3 cham xua't hien Ian t h u i " voi i = 1,2,3,4 K h i do: Bj la bie'n co " M a t 3 cham khong xua't hi?n Ian t h u i "
T a c o : A = B1.B2.B3.B4 uBj.B2.B3.B4 uB1.B2.B3.B4 uB,.B2.B3.B4 V| * S u y r a P ( A ) = P ( i ; ) p { B 2 ) P ( B 3 ) P( B 4 ) + P ( B i ) p ( B ^ ) p ( B 3 ) P ( B 4 ) V'> *
+ P ( B , ) P ( B 2 ) p ( i ; ) p ( B 4 ) + P ( B j ) P ( B 2 ) P ( B 3 ) p ( B ; )
M a P ( B j ) = ^ , P ( B , ) = ^ . D o d o : P U U 4 . -
6 324
V i d u 3. M p t h p p d u n g 4 vien bi xanh,3 vien bi do va 2 vien bi vang. Chon ngau nhien 2 vien b i :
1. T i n h xac suat de chpn dupe 2 vien bi cung mau. ;>-;, ; 2. T i n h xac suat de chpn dupe 2 vien bi khac mau.
G i a i . .. '
1. Gpi A la bie'n c6'"Chpn dupe 2 vien bi xanh"; B la bie'n c6'"Chpn dupe 2 vien b i do", C la bie'n co "Chpn dupe 2 vien bi vang" va X la bie'n co "Chpn dupe 2
vien b i cung m a u " . j , „ ,, Ta C O X = A B u C va cac bie'n co A, B,C doi m p t x u n g k h i c .
D o d o , taco: P(X) = P(A) + P(B) + P(C). • ^ , .
1...;;If',.;
V a y P ( X ) = l . ^ . ^ = A. ; 2, Bie'n C O "Chpn dupe 2 vien bi khac m a u " chinh la bie'n co'X .
Vay P(X) = 1 - P ( X ) = — . "r'-^ -o-inf: ic': .
18