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```5-3
Defn: Two events E1 and E2 are mutually exclusive if they cannot occur simultaneous
i.e. they have no common outcomes.
Ex.:
Roll a single die. Let E1 = {event of an even} and E2 = {event of an odd} then
E1 = {2,4,6} and E2 = {1,3,5}. E1 and E2 are mutually exclusive-no common
events.
Ex.
Select a single card from a standard deck:
Obtain a 7 or a jack. E1 ={ 7?,7?,7?,7? } and E2 = {J? ,J? ,J ? ,J? }. E1 and
E2 are mutually exclusive.
Obtain a club or a king A = {clubs }, and B ={ kings}. A and B are not
mutually exclusive since one King is also a club.
Get a face card or get an ace. A = {face card} and B = {ace}. Events A and B
are not mutually exclusive.
Thrm: (The Addition Rule)- If E1 and E 2 are events then
P(E1 or E2 ) = P(E 1 ) + P(E 2 ) - P(E 1 n E 2 ) .
E2
E1
Ee
Ex: 5-42 page 186
A store has 2 managers, 4 department heads, 16 clerks, 4 stock persons. (n = 26)
One person is selected at random. Find the probability that the person is a clerk
or a manager.
Solution:
P(clerk or manager) = 16 +
26
2 - 0 = 18 = .69
26
26
26
HW p 186 #38,39,40,43,48,49,52,57,61
#49 text A Stat class has 18jrs, 10srs, 6srs are Female, 12jrs are Male. Select one student
at random, find the probability it is a
1. jr or female
2. sr or female
3. jr or sr.
Solve: P(Jr or F) = P(Jr) + P(F) – P(Jr and F)
= 18 + 6 – 6
28 28 28
= 18
28
5-4 The Multiplication Rule
Defn: Events A and B are independent if A’s occurrence does not affect B’s occurance.
Otherwise they are dependent events.
Multiplication Rule 1: If A and B are independent events then
P(A and B) = P(A nB) = P(A)*P(B).
Examples.
1. Flip a coin and toss a die. Find P(Head and a 4).
P(Head and a 4 ) = P(H) * P(4) ,
since flipping the coin and tossing a die are independent events.
2. A card is drawn from a standard deck and replaced; draw a second and replace it.
Find the probability of drawing a Queen and an Ace.
P(Q and A) = P(Q) * P(A ) = 4 * 4
52 52
3. An urn contains 3 Red, 2 Blue, and 5 White balls. If two are drawn with replacement
then:
P(B and B) = P(two Blues) = 2 * 2
10 * 10
P(B and W) = 2 * 5
10 * 10
P(R and B) = 3* 2
10 * 10
Multiplication Rule 2
Defn: The conditional probability of event B given event A is the probability that B
occurs after event A has already occurred. (Notation P(B |A))
And the conditional probability of B given A is
P(B|A) = P(A and B) / P(A)
Ex. Draw 2 cards from a standard deck without replacement (dependent events)
Find the probability of drawing an Ace and then a King.
P(A and K) = P(A) * P(K |A) = 4 * 4 = 4
52 51 663
Note: P(K ?A) = probability of drawing a King (K) given that an Ace(A) has already
been drawn.
Ex. Given 25 microwave ovens, 2 are known to be defective. Select 2 at random
with-out replacement. Find the probability both are defective.
Solution:
Let D1 = event of a defective on the first draw. D2 = event of defective on 2nd draw.
P(D1 and D 2 ) = P(D1 ) * P(D1 /D2 ) = 2 * 1 = 1
25 24
300
Note P(D1 /D2 ) = probability of a defective on the second draw given that the first draw
was a defective.
Ex. Two cards or drawn without replacement. Find the following probabilities:
P(JJ) = P(J)*P(J ?J) = 4 * 3
52 51
P(CS) = P(S)*P(C ?S) = 13 * 13
52 51
P(CC) = P(C)*P(C ?C) = 13 * 12
52 51
J = Jack, C = Club, S = Spade
5-4 More Examples
Ex. 5-35 p 195
Draw four (4) cards from a standard deck replacing each card after its drawn. A win
secures if at least one ace is drawn. Find the probability of drawing a winning hand.
Sol: Let E = {event at least one Ace is drawn} then E = {event no ace is drawn}.
So
P (E) = 48 * 48 * 48 * 48 = 20,736
52 52 52 52
28,561
So P(E) = 1 –
P(E) = 1 – 20,736 = 0.27
28,561
Ex. 5-34 p194
Survey: Should women participate in combat? Select one respondent from the 100.
Yes
No
Total
Male
32
18
50
Female
8
42
50
Total
40
60
100
Find the probability the
(a) respondent answered Yes given the respondent was female.
(b) respondent was male given the respondent answered no.
Solve.
P(Y ?F) = P(Y and F) = 8/100 = 8
P(F)
50/100 50
P(M ?N) = P(M and N) = 18/100 = 18
P(N)
60/100 60
End
```
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