Rohan Chorge
Roll- 2013033
Group-5
Roll- 2013033
Group-5
Laws
of Probability
Multiplication
Law: If A1, · · · , Ak are independent
events, then
Pr(A1 ∩ A2 ∩ · · · ∩ Ak) = Pr(A1) Pr(A2)· ·
·Pr(Ak).
Addition
Law: If A and B are any events, then
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
Note: This law can be extended to more than
2 events.
Conditional
Law of Probablity-Dependent Event
The conditional probability of B given A
Pr(B|A) =
Pr(A ∩ B)/Pr(A)
A and B are independent events if and only
if
Pr(B|A) = Pr(B) = Pr(B|A)
'Addition Law example’
For
example, assume we wish to determine the probability of drawing a king and/or a
queen out of a deck of cards on only one draw. Using the addition rule for
probabilities, we get the following: P(King) = 4/52, P(Queen) = 4/52, and
P(King and Queen) = 0.
Since it is impossible to draw both a king and queen on the same draw, we can conclude that the probability of drawing either a king or queen from a deck of cards is 8/52
Since it is impossible to draw both a king and queen on the same draw, we can conclude that the probability of drawing either a king or queen from a deck of cards is 8/52
'Multiplication Law example’
Suppose we roll
one die followed by another and want to find the probability of rolling a 4 on the first die and
rolling an even number on the second die. Notice in this problem we are not
dealing with the sum of both dice. We are only dealing with the probability of 4 on one die only and then, as a
separate event, the probability of an even number on one die only.
P(4) = 1/6
P(even) = 3/6
P(even) = 3/6
So P(4
even) = (1/6)(3/6)
= 3/36 = 1/12
even) = (1/6)(3/6)
= 3/36 = 1/12
'Conditional
Law example’
If P(A)=.5, P(B)=.4, and P(AB)=.2 (hence P(AUB)=.7 and P(A'B')=.3),
P(A|B)=.2/.4=.5 and P(B|A)=.2/.5=.4.
