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Section 4

B depends on A ⟹p(B)



(B|A)P

P(B|A) = p(B) ⟹ B independent on A

Multiplicative rule (product rule):-

0 >B)= p(A)p(B|A); p(A)∩p(A ∴ ⟹. p(A) in two sides

()

p A B

P(B|A) =

where A &B can both occur.

Theorem:- A &B are independent events ⇔ p(A∩B)=p(A)p(B)

Example (1):- suppose that we have a fuse box containing 20 fuses of

which 5 defective. If 2 fuses are selected at random and removed from

the box in succession without replacing the first, what is the probability

that the both fuses are defective?

Solution:- probability that both fuse are defective ⟹ A(first event) ∩ B

(second event)

Without replacing the first ⟹ P(B|A) = p(B) ⟹ B independent on A

5 4 1

12 19 19

X =

B)=p(A)p(B)= ∩ p(A ∴

Example (2):- one bag contains 4 white balls and 3 black balls, and a

second bag contains 3 white balls and 5 black balls, one ball is drawn

from the first bag and placed unseen in the second bag. What is the

probability that a ball now drawn from the second bag is black?

black ball is drawn from the first bag. →

-Solution:

black ball is drawn from the second bag. →

white ball is drawn from the first bag. →

3 6 2

7 9 7

X =

= )

p()

)= p(

∩

∴ p(

4 5 20

7 9 63

X =

= )

p()

)= p(

∩

∴ p(

2 20 38

7 63 63

)}=

∩

)∪p(

∩

∴ p{p(

Example (3):- A small town has a fire engine and one ambulance

available for emergencies. The probability that the fire engine is

available when needed is 0.98, and the probability the ambulance is

available when called is 0.92 of the event of an injury resulting from a

burn building. Find the probability both the ambulance and the fire

engine will be available, assuming they operate independently?

Solution:- p(A(fire

engine)∩B(ambulance))=p(A)p(B)=0.98*0.92=0.9016.

Example (3):- An electrical system consists of 4-compenent. The system

works if component A and B work and either of the component C or D

works. The reliability(the probability of working) of each component.

Find the probability that

a) the entire system work.

b) the component C dose not work given that the entire system work.

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Section 4 P(B|A)  p(B) ⟹ B depends on A P(B|A) = p(B) ⟹ B independent on A Multiplicative rule (product rule):P(B|A) = p (A B ) . p(A) in two sides ⟹ ∴p(A ∩B)= p(A)p(B|A); p(A)>0 p (A ) where A &B can both occur. Theorem:- A &B are independent events ⇔ p(A∩B)=p(A)p(B) Example (1):- suppose that we have a fuse box containing 20 fuses of which 5 defective. If 2 fuses are selected at random and removed from the box in succession without replacing the first, what is the probability that the both fuses are defective? Solution:- probability that both fuse are defective ⟹ A(first event) ∩ B (second event) Without replacing the first ⟹ P(B|A) = p(B) ⟹ B independent on A ∴ p(A ∩ B)=p(A)p(B)= 5 4 1 X = . 12 19 19 Example (2):- one bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls, one ball is drawn from the ...
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