Example
In how many ways can a supermarket manager display 5 brands of cereals in 3 spaces on a shelf?
The number of ways is:
P(A)=(5!)/((5-3)!)=(5!)/(2!)=60
6)A combination. The number of combinations of n distinct objects taken r at a time.
With a combination we still select r objects from a total of n, but the order is no longer considered. With a combination we still select r objects from a total of n, but the order is no longer considered.
The number of combinations of n objects taken r at a time is given by the formula:
C(n,k) = n!/[k!(n - k)!]
An Example of Combinations
Now we will answer the following question: how many combinations are there of two letters from the set {a,b,c}? Since we are dealing with combinations, we no longer care about the order. We can solve this problem by looking back at the permutations, and then eliminating those that include the same letters. As combinations, ab and ba are regarded as the same. Thus there are only three combinations: ab, ac, bc.
7. Additive Rules (Addition formula of probability). 1) a1, a2, . . . , An are mutually exclusive. 2) a1, a2, . . . , An eny events.
If A and B are events, the probability of obtaining either of them is:
P(A∩B) = P(A) + P(B) - P(A U B)
If the events A1, A2, . . . , An are mutually exclusive( that is, if events cannot occur simultaneously), the last term [P(A1, A2, . . . , An)] will be 0. Thus the addition rule with mutually exclusive events becomes:
P(A1∩ A2∩ . . . ∩ An) = P(A1) + P(A2)+…+P(An)
Example:
Suppose a high school consists of 25% juniors, 15% seniors, and the remaining 60% is students of other grades. The relative frequency of students who are either juniors and seniors is 40%. We can add the relative frequencies of juniors and seniors because no student can be both junior and senior.
P(J or S) = 0.25 + 0.15 which equals 0.40
Generalized Addition Rule for Any Events
The above formula can be generalized for situations where events may not necessarily be mutually exclusive. For any events A1, A2, . . . , An, the probability of A1∩ A2∩ . . . ∩ An is the sum of the probabilities of A1, A2, . . . , An minus the shared probability of A1U A2 U. . . UAn:
P(A ∩ B) = P(A) + P(B) - P(A U B)
8. Conditional probability.
Let A and B
P(B|A)=
=
*
P(B|A)
The probability of 2 events will happened equals the probability of first event will happened multiplied the probability of event B will happened gives the first event A has already happened.
Theorem on
conditional probability. For any free events A1,A2,A3
we
have P(A1
A2
A3)=
P(A1)*
P(A2
A1)*P(A3|A1
A2)
Some theorems on probability.
1)If
A1
A2
P(A1)
P(A2)
and P(A1
A2)
= P(A1)-
P(A2)
2)For event
A: 0
P(A)
1
3)
-
emptyset P(
)
4)If Ac- complement of A P(Ac)= 1 –P(A)
5)If A=
A1
A2
A3
An
P(A)=
P(A1)+
P(A2)+
…+P(An)
In
particular A=
P(A1)+ P(A2)+ …+P(An)=1
6)A,B- any
events P(A
)=P(A)+P(B)-P(A
B)
7)A1,A2,A3
P(A1 A2 A3)= P(A1)+ P(A2)+ P(A3)-P(A1 A2)-P(A2 A3)-P(A1 A3)+ P(A1 A2 A3)
8)If an event A must result in the happened of one of mutually exclusive event A1 A2 A3 An then the probability of a : P(A)= P(A A1)+ P(A A2)+…+ P(A An)
Example. The square is enter a circle. The point at random gets to a circle. Find the probability that a point will get to a square.
A-Event the
point gets to the square. Solution S(G)=
R2
S(g)=2R2
P(A)=