Download Lecture # 3 Counting Techniques 1. Multiplication Principle: Let E

Lecture # 3 Counting Techniques
1. Multiplication Principle:
Let E and F be two sets of outcomes of an experiment. We use the multiplication rule to calculate the
probability that both E and F occur.
Definition: Let E1, E2, E3, ..., En be a sequence of n independent events. If Ek can occur Mk ways for k =
1, 2, 3, …, n, then there are M1, M2, M3, …, Mn ways for all events to occur.
Examples:
Example 1) How many different combinations can be made for a briefcase that has a 3-dial lock, with
each dial having numbers from 0 – 9 available?
Answer: A briefcase 3-dial lock has possibly 3 slots with numbers.
0
0
0
Pick a no.
Pick a no.
Pick a no.
for Dial 1
for Dial 2
for Dial 3
E1
E2
E3
Where E1 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} => n(E1) = 10
E2 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} => n(E2) = 10
E3 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} => n(E3) = 10
This means, there are 10 choices for dial 1, 10 for dial 2 and 10 for dial 3. Therefore, there are 10 x 10 x
10 = 1000 possible combinations of 3 digit numbers to unlock the briefcase.
Intuitively, we are making the combinations such as 0-0-0, 0-0-1, 0-0-2, …, 9-9-8, 9-9-9. When we
count all these combinations, they are 1000 in number.
Example 2) How many license plates on car can be made if the first 3 entries must be letters, followed
by 3 numbers so that you will be able to uniquely identify a car?
Answer: The idea is to list all possible license plates that uniquely identify any car. With 3 letters at the
first place and 3 digits on the remaining 3 places, the license plate will look like:
A
B
A
2
1
8
Pick a letter
for entry 1
26
Pick a letter
for entry 2
26
Pick a letter
for entry 3
26
Pick a number
for entry 4
10
Pick a number
for entry 5
10
Pick a number
for entry 6
10
The number of choices for the first entry = 26 (there are total 26 English letters for A – Z)
The number of choices for the second entry = 26
The number of choices for the third entry = 26
The number of choices for the fourth entry = 10 (1 digit number can be from 0 – 9)
The number of choices for the fifth entry = 10
The number of choices for the sixth entry = 10
Using the multiplication rule, we have the total number of possible license plates that can be made = 26
x 26 x 26 x 10 x 10 x 10 = 17, 576, 000
Example 3) Suppose you take a multiple choice exam that has 10 questions, each question has 5
answers. How many different ways could the exam be answered?
Instructor: Ms. Azmat Nafees
8-Sep-09
Answer:
Let’s say the five choices for each question can be A, B, C, D, & E
Then all questions from 1 – 10 will have choices:
1.
A, B, C, D, E
2.
A, B, C, D, E
.
…..
.
…..
.
…..
.
…..
8.
A, B, C, D, E
9.
A, B, C, D, E
10.
A, B, C, D, E
If an event for Q.1 could be = E1 = {guessing an answer for question 1} => n(E1) = 5
An event for Q.2 could be = E1 = {guessing an answer for question 2} => n(E2) = 5, and so on.
Then the total number of ways the exam could be answered = 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 = (5)10
= 9, 765, 625
Example 4) Two cards are drawn without replacement from a deck of 52 cards in such a way that:
A1 = an ace on the first draw
A2 = an ace on the second draw
Total number of ordered two cards= 52 x 51
Total number of two cards= 4 x 3
Total number of 1st card ace, 2nd not ace = 4 x 48
Total number of 1st card not ace, 2nd ace = 48 x 4
Total number of both cards not ace = 48 x 7
As, Total cards = (Ace) x (Not Ace)
52 = 4 x 48
Example 5) A box contains 100 microchips produced by factory 1 and 2. Some are defective and some
are non-defective (good).
Defective (A)
Factory 1 (B)
15 (A ∩ B)
Factory2 (B’)
5 (A ∩ B’)
Total
20
45 (A’ ∩ B)
35 (A’ ∩ B’)
80
40
100
Good (A’)
60
Total
a) P(A) = n(A)/n(S) = 20/100
b) P(A’) = 80/100
c) P(B’) = 60/10
d) P(B’) = 40/100
e) P(A ∩ B) = 15/100
f) P(A’ ∩ B’) = 35/100
g) P(A ∩ B’) = 5/100
Instructor: Ms. Azmat Nafees
8-Sep-09
2. Multiplication theorem (for without replacement):
P(A∩B) = P(A) . P (B/A) = P(B). P(A/B)
For instance, in previous example no. 5 of microchip we had:
P(A∩B) = 15/100
We can compute the same probability such as P(A ∩ B) = P(B) x P(A/B) = (20/100) x (15/20) = 15/100
Similarly in example 4, if 5 cards are drawn from a deck of 52 cards, then possible outcomes:
With replacement: 52 x 52 x 52 x 52 x 52 = (52)ହ
Without replacement: 52 x 51 x 50 x 49 x 48
3. Conditional Probability:
The probability of an event occurring given that another event has already occurred is called a
conditional probability. The probability that event B occurs, given that event A has already occurred is
P(B|A) = P(A and B) / P(A)
This formula comes from the general multiplication rule.
Example: Using the joint probability table in question 5:
Probability of a defective chip while it has been determined that a micro chip was produced in factory 1
P(A/B) = P(A ∩ B)/P(B) = 15/100 / 60/100 = 15/60
Probability of a good (A’) microchip, if it is determined that chip was produced in factory 2 (B’)
P(A’/B’) = 35/40
Or, P(A’/B’) = P(A’ ∩ B’)/P(B’) = (35/100) /(40/100) = 35/40
Probability of a defective (A) microchip given that it is produced in factory 2 (B’)
P(A/B’) = 5/40
Or P(A/B’) = P(A ∩ B’)/P(B’) = (5/100) /(40/100) = 5/40
Example 6) Two cards are drawn without replacement from a deck of cards.
Let ‫ܣ‬ଵ = an ace on the first draw, and ‫ܣ‬ଶ = an ace on the second draw
Partitioning the number of ways to draw 2 cards:
Ace(࡭૛ )
Not Ace (࡭′૛ )
Total
࡭૚ (Ace)
4x3
4 x 48
4 x 51 = n(‫ܣ‬ଵ )
࡭૚ ′ (Not Ace)
48 x 4
48 x 47
48 x 51 = n(‫ܣ‬ଵ ′)
Probability of getting ‫ܣ‬ଵ ܽ݊݀ ‫ܣ‬ଶ = P(‫ܣ‬ଵ ∩ ‫ܣ‬ଶ ) =
Instructor: Ms. Azmat Nafees
௡(஺భ ∩஺మ )
௡(ௌ)
Total
4 x 51 = n((‫ܣ‬ଶ )
48 x 51 = n(‫ܣ‬′ଶ )
52 x 51 = n(S)
(ସ×ଷ)
= (ହଶ×ହଵ)
8-Sep-09
P(‫ܣ‬ଵ ) regardless of whatever comes on the 2nd draw will be:
(4 x 51) / (52x51) = 4/52 = 1/13
Similarly, P(‫ܣ‬ଶ ) regardless of whatever will be on 1st draw”
(4 x51) /(52x51) = 4/52 = 1/13
The conditional probability that one ace is drawn on the 2nd draw given that an ace was obtained on the
1st draw:
P(‫ܣ‬ଶ /‫ܣ‬ଵ ) =
௉(஺భ ∩ ஺మ )
௉(஺భ )
=
(ସ×ଷ)∕(ହଶ×ହଵ)
ସ∕ହଶ
=
ଵ
ଵ଻
Or P(‫ܣ‬ଶ /‫ܣ‬ଵ ) = (4 x 3) /(4 x 51) = 3/51 = 1/17 (computing directly from the joint probabilities’ table)
Now the probability of ‫ܣ‬ଵ ܽ݊݀ ‫ܣ‬ଶ without replacement is:
ܲ(‫ܣ‬ଵ ∩ ‫ܣ‬ଶ ) = ܲ(‫ܣ‬ଵ ). ܲ(‫ܣ‬ଶ /‫ܣ‬ଵ )
= (4/52) · (3/51) = (4 x 3)/(52 x 51) = 1/(13 x 17)
Properties of conditional probability:
1) 0 ≤ P(A|B) ≤ 1
2) If ‫ܣ‬ଵ ܽ݊݀ ‫ܣ‬ଶ are mutually exclusive events, then:
P(‫ܣ‬ଵ ∪ ‫ܣ‬ଶ /B) = P(‫ܣ‬ଵ /‫ )ܤ‬+ ܲ(‫ܣ‬ଶ /‫)ܤ‬
L.H.S. P(‫ܣ‬ଵ ∪ ‫ܣ‬ଶ /B)
Since P(AUB) = P(A) + P(B) – P(A∩B)
Therefore, P(‫ܣ‬ଵ ∪ ‫ܣ‬ଶ /B) = P (‫ܣ‬ଵ /‫ )ܤ‬+ ܲ(‫ܣ‬ଶ /‫ – )ܤ‬P(‫ܣ‬ଵ /࡮ ∩ ‫ܣ‬ଶ /B)
This is true for not mutually exclusive events.
3) P(‫ܣ‬ଵ ∩ ‫ܣ‬ଶ /B) = P(‫ܣ‬ଵ /‫ )ܤ‬+ ܲ(‫ܣ‬ଶ /‫ – )ܤ‬P(‫ܣ‬ଵ /࡮ ∩‫ܣ‬ଶ /B) if ‫ܣ‬ଵ ܽ݊݀ ‫ܣ‬ଶ are not mutually exclusive
events.
4) P(A/B) = 1 – P(A’/B)
Proof:
We know P(AUA’/B) = P(S)
[as AUA’ = S => P(AUA’) = P(S)]
⇒ P(A/B U A’/B) = 1
[ as P(AUA’) = 1 => P(A) + P(A’) = 1]
⇒ P(A/B) + P(A’/B) = 1
⇒ P(A/B) = 1 – P(A’/B)
Or P(A’/B) = 1 – P(A/B)
e.g the sample of two coins tossed = S = {HH, HT, TH, TT},
if A = {HH} then A’ = { HT, TH, TT}
Therefore, AUA’ = S and A∩A’= { }
Instructor: Ms. Azmat Nafees
8-Sep-09
Exercise (selective questions from text book):
Q23. Let an event A = color TV set is on, and B = Black and White TV set is on.
We are given that P(A) = 0.4, P(B) = 0.3, and P(AUB) = 0.5
a) P(both are on) = P(A B)
As P(AUB) = P(A) + P(B) – P(A∩B)
0.5 = 0.4 + 0.3 – P(A∩B)
P(A∩B) = 0.4 + 0.3 – 0.5 = 0.2
P(A∩B) = 0.2
These probabilities can be expressed in form of a joint probability table:
A(Color TV on) A’(Color TV off)
0.1 (A’∩B)
0.2 (A∩B)
B(B&W TV on)
0.2 (A∩B’)
0.5 (A’∩B’)
B’(B&W TV off)
Total
0.4
0.6
b) P(A∩B’)= 0.2
c) P(exactly one is on) = 0.2 + 0.1 = 0.3
d) P(A’∩B’) = 0.5
Total
0.3
0.7
1
Q22. In this statement, we are given that:
P(winning 1st race) = 0.7 = P(A)
P(winning 2nd race) = 0.6 = P(A)
P(winning both race) = 0.5 = P(A ∩ B)
A
A’
Total
0.5
0.1
0.6
B
0.2
0.2
0.4
B’
0.7
0.3
1
Total
a) P(at least 1 race win) = 1 – P(neither race win) = 1 – 0.2 = 0.8
b) P(Exactly one race win) = 0.2 + 0.1 = 0.3
c) 0.2 = P(neither race win)
Q25. In this statement, we are given that: B + G = T => 2 + 3 = 5
Let’s say two players, i.e. Player A and Player B
a) P(A good) = 3/5
b) P(B good / A good) = 2/4 = 1/2
c) P(B good / A bad) = 3/4
d) P(B good A good) = P(A good) . P(B good / A good)
= (3/5) . (2/4)
= 3/10
e) P(B good) = 3/5
f) P(A good / B good) = 2/4 = ½
As, B + G = T
2+3=5
2+2=4
Instructor: Ms. Azmat Nafees
[for without replacement or conditional probability]
8-Sep-09