Probability theory.pptx

Probability
theory
Eng. Ahmed Yahiya Barakat

Experiments &
Sample
Spaces
1. Experiment
• Process of observation that leads to a single outcome
that cannot be predicted with certainty
2. Sample point
• Most basic outcome of an
experiment
3. Sample space (S)
• Collection of all possible outcomes
Sample Space
Depends on
Experimenter!

Sample Space Properties
Experiment: Observe Gender
© 1984-1994 T/Maker Co.
1. Mutually Exclusive
2 outcomes can not occur
at the same time
— Male & Female in
same person
2. Collectively Exhaustive
One outcome in sample
space must occur.
— Male or Female

Examples
1. Tossing a coin – outcomes S ={Head, Tail}
2. Rolling a die – outcomes
S ={ , , , , , }
={1, 2, 3, 4, 5, 6}

S
HH
TT
TH
HT
Sample Space S = {HH, HT, TH, TT}
Venn Diagram
Outcome
Experiment: Toss 2 Coins. Note Faces.
Compound
Event: At
least one
Tail

Item 1 2 3 4 5 6
1 1/36 1/36 1/36 1/36 1/36 1/36
2 1/36 1/36 1/36 1/36 1/36 1/36
3 1/36 1/36 1/36 1/36 1/36 1/36
4 1/36 1/36 1/36 1/36 1/36 1/36
5 1/36 1/36 1/36 1/36 1/36 1/36
6 1/36 1/36 1/36 1/36 1/36 1/36
Sum two (1,1) 1/36
Sum three is (2,1), (1,2) 2/36
Sum four is (2,2), (3,1), (1,3) 3/36
Sum five is (2,3), (3,2), (4,1), (1,4) 4/36
Sum six is (3,3) ,(5,1), (1,5), (4,2), (2,4) 5/36
Sum seven (5,2), (2,5), (3,4), (4,3), (6,1), (1,6) 6/36
Sum eight (5,3), (3,5), (4,4), (6,2), (2,6) 5/6
Sum nine (5,4), (4,5), (6,3), (3,6) 4/6
Sum ten (5,5), (6,4), (4,6) 3/6
Sum eleven (5,6), (6,5) 2/6
Sum twelve (6,6) 1/6
Sum less than 7= 1/36+2/36+3/36+4/36+5/36=0.417
Sum equal to 10= 3/6= 0.333
The two events are mutually exclusive (you can’t get
the two events in the same two rolls)

Event A >>>> P(x<5) = P(x=1)+ P(x=2)+ P(x=3) + P(x=4)= 0.4
Event B >>>> P(x= odd)= P(x=1)+ P(x=3)+ P(x=5) + P(x=7) + P(x=9)= 0.5
P A or B = P(A υ B) = P(A) + P(B)= 0.9
P(A’)= 1-P(A)= 0.63
P(AՈB)= P(A).P(B)= 0.37*0.44= 0.163
P(AՈB’)= P(A).(1-P(B))= 0.37*(1-0.44)= 0.207
P(A’ՈB’)= 0.63*0.56= 0.353

B
A
P(A Ս B) = P(A) + P(B)= 0.75 >> P(B) = 0.75
P(A’ Ո B) = (1-P(A)) . P(B)= 5/8 >> P(A) = 0.167

P(A)= 0.6
P(B)= 0.5
P(AՈB) = 0.6*0.5= 0.3
𝑃 𝐴′
𝐵′
=
𝑃(𝐴′
Ո𝐵′
)
𝑃(𝐵′)
=
0.4 × 0.5
0.5
= 0.4

The entire possibilities for the chosen person= 4C2=6
The favorable outcome= 2C2= 1
P(A)= 1/6

Probability theory.pptx

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