# Seventh Lecture ```Previous Lecture:
We learnt the following:
 Random Experiment.
 Sample Space.
 An Event.
 The Probability.
This lecture we will learn
 Conditional Probability.
 Independent Events.
 Multiplication Rules.
Conditional Probability
 We often want to find the probability of an event A
occurring given that B has occurred.
 In these cases, we are conditioning on B, and we write
P(A|B).
 B serves as a new (reduced) sample space.
 P(A|B) is that fraction of P(B) which corresponding to
P(A ∩ B).
Definition:
The conditional probability of B, given A, denoted by
P(A|B) is defined by
Where P(B) &gt; 0.
Example(conditional Probability):
 Throw two dice. The sum shown is 10. What is the
probability that one 6 is cast?
Solution:
 Let A be the event “ one 6 is cast”.
 Let B the event “the sum equals 10”.
 We want to calculate P(A|B).
 A={(1,6 ) , (2,6), (3,6 ) ,(4,6), (5,6) ,(6,6) , (6,1) , (6,2),
(6,3) ,(6,4), ( 6,5)}.
 B= {(6,4), (4,6), (5,5)}.
Example cont.
P (A ∩ B) = 2/36.
P(B) = 3/36.
And we know
So:
P(A|B)= 2/3.
Example
If we have P(A)=0.35, P(B)=0.5, P(A⋃B)=0.8, Find
P(A⋂B), P(A|B), P(B|A)
P(A⋂B)=P(A)+P(B)-P(A⋃B)
0.35+0.5-0.8=0.05
P(A|B)=
𝑃(𝐴⋂𝐵)
𝑃(𝐵)
0.05/0.5
P(B|A)=
𝑃(𝐴⋂𝐵)
𝑃(𝐴)
Example
in a class of 30 students ,there are 17 girls and 13 boys .A
total of 5 students made an A grade on their report card ,
three of which were girls .if a student is a chosen at
random . The probability of getting a girl or an A
student is ……..
Independence
 A and B are independent if knowledge that one event
has occurred does not change the probability that the
other will occur.
 Definition:
Two events A and B are independent if and only if
P(A|B)= P(A) or P(B|A)=P(B)
Provided the existences of the conditional probability.
Otherwise, A and B are dependent.
Example:
Classifying Events as Independent or dependent:
 Tossing a coin and getting a head(A), and then rolling
a die and obtaining a 6 (B).
 Driving over 85 miles per hour (A), and then getting in
a car accident(B).
 Exercising frequently (A) and having a low blood
pressure.
Solutions:
 P(B|A)= 1/6 and P(B)= 1/6. The occurrence of A does
not change the probability of occurrence of B, so the
events are independent.
 If you drive over 85 miles per hour, the chances of
getting in car accident are greatly increased, so these
events are dependent.
 If you exercise regularly it will affect the blood pressure
by make it lower, so these events are dependent.
Example
 Let A and B are independent events as follows:
P(A)= 0.5, and P(B)=0.6, Find P(A|B), P(B|A).
Solution:
 Because the event are independent
 P(A|B)=P(A)= 0.5, and P(B|A)=P(B)=0.6.
Multiplicative Rules
Theorem:
If in an experiment the events A and B can both occur,
then :
P (A ∩ B)= P(A)P(B|A) ,
provided
that P(A) &gt;0.
Example:
Two cards are selected, without replacing the first card
from a standard deck. Find the probability of selecting
a king and then selecting a queen.
Solution:
Because the first card not replaced, the events are
dependent
P (K ∩ Q)= P(K)P(Q|K)
=4/52 . 4/51
=16/2652 ≈ 0.006.
Theorem:
Two events A and B are independent if and only if
P (A ∩ B)= P(A)P(B) .
Therefore, to obtain the probability that two
independent events both occur, we simply find the
product of their individual probabilities.
Example:
A coin is tossed and a die is rolled. Find the probability
of getting a head and then rolling a 6.
Solution:
The events are independent.
P(A∩ B)= P(A). P(B)
=1/2.1/6=1/12≈ 0.083
So the probability of tossing a head and then rolling a
Example
A box contains 5 white balls, and 7 blue balls. If two balls were
randomly chosen (one after another), calculate the following
probabilities:
-The two balls are white, chosen with replacement.
-The two balls are blue, chosen without replacement.
Example
Let A and B be two events, with probabilities P(A) = 0.6,
and P(B) =0.2
Calculate the following probabilities: P(A`), P(A⋂B),
P(A⋃B), P(A|B), P(B|A), if:
A and B are mutually exclusive (disjoint). •
A and B are independent. •
Example
 The probability that a salmon swims successfully
through a dam is 0.85. Find the probability that three
salmon swim successfully throw the dam.
 Solution
The probability that each salmon is successful is 0.85,
one salmon's chance of success is independent of the
others chances
P(3 salmon successful)= (0.85). (0.85).(0.85)
≈0.614
Example
 Find the probability that none of the three salmon is
successful.
 Solution
Because the probability of success for one salmon is 0.85,
the probability of failure for one salmon is 1 -0.85=0.15
P(none of the three is successful) = (0.15).(0.15).(0.15)
≈0.003
So the probability of none are successful is about 0.003.
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