Probablity
experiment : Pareekshanam
Random experiment: kittan sadyada ulladu we know.
if a coin tossed head and tail will be the result.
if a coin tossed ={H,T}
if a die tossed:{1,2,3,4,5,6}
if a exam:{P,F}
sample space: set of All possible outcomes "S"
S={H,T}
S={1,2,3,4,5,6}
S{P,F}
Event:Subset of sample space "C" contains (notation for subset)
A={H} =ACS
B={123} =BCS
C={P} =CCS
If a coin toosed then D={H,T} so event is equal subset DS
D={} null set (impossible event) sample space with null space is (null set)
A={H,T} varam Sure event (full outcomes) (sample space with full outcomes is sure event)
Type of events
1.Mutually exclusive event
2.Exhaustive event
3. Independent event
Mutually exclusive
2 events A and B (IF they do not occur together)
Example tose a coin S={H,T}
A={H}
B={T}
A kittiyal B yae thadanju . B kittiyal A thadanju (A occurance prevents the B)(B occurance prevents A)
Kasera kali is a mutualiy exclusive
There for
A∩B=ɸ (null set) (cant see any common elements )
Independant event
A and B 2 events shooting a target if A shooted target because of B can shoot or not. If an event occur will not effect other event. so it is called independent event
Exhaustive Event
A={1,2,3}
B={4}
C={5,6}
A∪B∪C={1,2,3,4,5,6} =S
If we take union of events then we get sample space (All possible out comes)
is called exhaustive event.
Probablity
Probablity always for an event
P(event)
P(A)=n(A)/n(s) (no of cases favorable to A )/ (Total number of cases)
A S={1,2,3,4,5,6}
A={1,3,5}
B={2,4,6}
C={3,6}
D={2,3,5}
P(A)=n(A)/n(S) (number of elements of a by number of sample space)
P(A)=3/6=1/2
P(B)=3/6=1/2
P(C)=2/6
P(D)=3/6=1/2
Properties
i) P(A)>=0 Always posetive
ii)0<=p(A)<=1
iii)p(S)=1; P(S)=n(S)/n(S)=6/6=1
iv)p(ɸ)=0; p(ɸ)=n(ɸ)/n(S)=0/6=0
Compliment of an event
C={3,6} D={2,3,5}
C|={1,2,4,5} (Complimen of C) D|={1,4,6} (Compliment of D)
P(C)=2/6 ; P(C|)=4/6 P(D)=3/6; P(D|)=3/6
P(C)+P(C|)=2/6+4/6=6/6=1 P(D)+P(D|)=3/6+3/6=6/6=1
P(A)+P(A|)=1
Mutually Exclusive
A⋂B=ɸ(reffer above)
there for P(A⋂B)=p(ɸ)=0
Independant event
P(A⋂B)=p(A)p(B)
Can sample Space
A coin is tossed
S={H, T}
If 2 coins are tossed
S={HH,HT,TH,TT}
if n coins are toosses n(S)=2raiseton(here 2 is head and tail)\
so if 3 coins were tossed n(S)=2cube=2*2*2=8
If 3 coins tossed
S={HHH,HHT,HTH,THH,HTT,THT,TTH}
number of sample space if a die throw
6^n
S=if one die={1,2,3,4,5,6}
6^1=6
if 2 die throw 6^2=6*6=36
S={(1,2)(1,3)(1,4),(1,5)(1,6)(2,1)(2,2).......(6,6)}
Card 52 total (13 num 4 groups)
♠ ♣ ❤ ❖
spade clubs hearts diamond
13 13 13 13
black card =26
red card =26
K
Q Picture card is first 3
J
A
2
3
4
5
6
7
8
9
10
black king =2
face card =4 in each =36
Picture card =3 in each=3*4=12
HOW MANY RED PICTURE CARD? 3*2=6
NUMBER CARD =9 *4=36
HOW Many black number card=9*2=18
Addition Theorem
P(AUB)=P(A)+P(B)-P(A⋂B) ( union is or and intersection is AND)
mutually exclusive
A and B will not happen together
P(A⋂B)=0
so
P(AUB)=P(A)+P(B)
Independant event
P(AUB)=P(A)+P(B)-P(A⋂B)
here in independant
P(A⋂B) =P(A)P(B)
P(AUB)=P(A)+P(B)-P(A)P(B)
Addition Theorem
P(AUB)=P(A)+P(B)-P(A⋂B)
A∩B'
A'∩B
Demorgans Law
(AUB)'=A'∩B'
(A∩B)'=A'UB'
P(AUBUC)=P(A)+(B)+P(C)-P(A∩B)-P(B∩C)-P(A∩C)+P(A∩B∩C)
Add single
Substrat double
add all
Condidtional Probablity
P(A/B)=P(A∩B)/P(B)
P(B/A)=P(A∩B)/P(A)
Questions
i)A total of 9
ii) not getting a doublet
S={(1,1).........(6,6)}
i)A={a total of 9}
A={(3,6)(4,5)(5,4)(6,3)}
P(A)=n(A)/n(S)
P(A)=4/36
P(A)=1/9
ii) not getting a doublet
here doublets are={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}=total 6
P(not getting a doublet)=30/36=5/6
2. Two dies are thrown
i) A total of 8
{(2,6)(3,5)(4,4)(5,3)(6,2)}
p(A)=5/36
ii) B a doublet
B={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}=6/36=1/6
(iii) Atotal of 9 or 11 (OR) (AUB)
C={A total of 9 or 11} for union ellam ezhudanam
={(3,6)(4,5)(5,4)(6,3)(5,6)(6,5)}=6/36=1/6
iV) Both odd digits
A={(1,1)(1,3)(1,5)(3,1)(3,3)(3,4)(5,1)(5,3)(5,5)}=9/36
A=1/4
V) even num as sum
A={(1,1)(1,3)(1,5)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(4,2)(4,4)(4,6)(5,1)(5,3)(5,5)(6,2)(6,4)(6,6)}
=18/36=1/2
2. A box contains 5 red, 6 green and 4 white balls. A ball is drawn at random what is the probablity that the ball drawn is
i)red
ii)green
iii)either red or green
iv)not red
5 R.
6G
4 W 1 ball taken
i)p(red ball)=5/15=1/3
ii)P(green ball) =6/15=2/5
iii)either red or green 5 + 6/15=11/15
iv)not red=6+4/15=10/15=2/3
3.A bag contains 6 red, 5 white and 4 black balls. Two balls are drawn
6 R
5 W
4 B 2 ball drwan
i) both are red
P(R)=
select 2 or more from group is called combination
here 3 groups 3C2
i)if both are red
6 C 2/15C2 6C2=6*5/1*2
6C3=6*5*4/1*2*3
6*5/1*2divided by 15*14/1*2
3*5/15*7
15/105=3/21=1/7
ii) Both are white
5C2/15C2=5*4/1*2/15*14/1*2=
5*2/15*7=2/21
iii)Probability none is red
9C2/15C2=9*8/1*2/15*14/1*2=9*4/15*7=36/105
iV) Probablity of(1 white and 1 black )
5C1 and 1 black And anengil (*) or anengil (+)
5c1*4C1/15C2
=5*4/105=20/105=4/21
4. 8 Red ,5 White and 7 Blue balls
P(1 white and 2 Blue balls)
5C1*7C2=
5C1=5/1=5
7C2=7*6/1*2=7*3=21
5*21/20C3= 20*19*18/1*2*3=10*19*18=10*19*6=19*6=1140
105/1140
P(1 White ,1 Red and 1 blue)
5C1*8c1*7C1=5*8*7/20C3=280/1140
P(None is Red)
5+7C3/20C3=12*11*10/1*2*3/1140=220/1140
mukalil cherudum thazhaey valadum for all probablity
Are all of same colour
P(3 red or 3 white or 3 blue_
8c3+5c3+7c3/20c3
=8*7*6/1*2*3=4*7*3=84
=5*4*3/1*2*3=5*2=10
=7*6*5/1*2*3=7*3*5/1*3=105
84+10+105/1140=199/1140
56+10+35/1140
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