HOA1&2 - Module 3 - PREHISTORCI ARCHITECTURE OF KERALA.pptx
Chapter 4-Probability-2ndpart.ppt
1. EBE 101
NAME OF SCHOOL
FACULTY OF ENGINEERING AND INFORMATION
TECHNOLOGY
PREPARED BY:
KALAIVANI A.TARUMARAJA
CHAPTER 4
ENGINEERING MATHEMATICS I
PROBABILITY
2. Slide 2 of 14
TOPIC
Complements, Unions, & Intersections
Suppose A & B are events.
3. Slide 3 of 14
TOPIC
The complement of A is everything in the sample space S that is NOT in A.
A
S
If the rectangular box
is S, and the white
circle is A, then
everything in the box
that’s outside the circle
is Ac , which is the
complement of A.
P(A) =0.4
P(Ac) = 1- 0.4 = 0.6
4. Slide 4 of 14
TOPIC
Theorem
Pr (Ac) = 1 - Pr (A)
Example:
If A is the event that a randomly selected
student is male, and the probability of A is
0.6, what is Ac and what is its probability?
Ac is the event that a randomly selected
student is female, and its probability is 0.4.
5. Slide 5 of 14
TOPIC
The union of A & B (denoted A U B) is everything in the
sample space that is in either A or B or both.
A
S
The union of A & B is the whole white area.
B
6. Slide 6 of 14
TOPIC
The intersection of A & B (denoted A∩B) is everything in
the sample space that is in both A & B.
S
The intersection of A & B is
the pink overlapping area.
B
A
7. Slide 7 of 14
TOPIC
Example
Adrina & Steven are planning for trip at
CANADA & SINGAPORE .
Suppose Canada (C) & DUBAI (D) are equally
likely.
What is the sample space S?
S = {CC, DD, CD, DC}
8. Slide 8 of 14
TOPIC
Example continued
If E is the event that both COUNTRY are
the same CHOICE, what does E look like
& what is its probability?
E = {CC, DD}
Since Canada & Dubai are equally likely,
each of the four outcomes in the sample
space S = {CC, DD, CD, DC} is equally
likely.
So, Pr(E) = 2/4 = 1/2 = 0.5
9. Slide 9 of 14
TOPIC
Example cont’d: Recall that E = {CC, DD} & Pr(E)=0.5
What is the complement of E and what is its
probability?
Ec = {CD, DC}
Pr (Ec) = 1- Pr(E) = 1 - 0.5 = 0.5
10. Slide 10 of 14
TOPIC
Example continued
If F is the event that at least one of the
COUNTRY is a DUBAI, what does F look
like & what is its probability?
F = {CD, DC, DD}
Pr(F) = 3/4 = 0.75
11. Slide 11 of 14
TOPIC
Recall: E = {CC, DD} & Pr(E)=0.5
F = {CD, DC, DD} & Pr(F) = 0.75
What is E∩F?
{DD}
What is its probability?
1/4 = 0.25
12. Slide 12 of 14
TOPIC
Recall: E = {CC, DD} & Pr(E)=0.5
F = {CD, DC, DD} & Pr(F) = 0.75
What is the EUF?
{CC, DD, CD, DC} = S
What is the probability of EUF?
1
If you add the separate probabilities of E & F together,
do you get Pr(EUF)? Let’s try it.
Pr(E) + Pr(F) = 0.5 + 0.75 = 1.25 ≠ 1 = Pr (EUF)
Because of (DD) repeat twice.
13. Slide 13 of 14
TOPIC
A formula for Pr(EUF)
Pr(EUF) = Pr(E) + Pr(F) - Pr(E∩F)
If E & F do not overlap, then the intersection is
the empty set, & the probability of the
intersection is zero.
When there is no overlap,
Pr(EUF) = Pr(E) + Pr(F).
14. Slide 14 of 14
TOPIC
Conditional Probability of A given B
Pr(A|B)
Pr(A|B) = Pr (A∩B) / Pr(B)
15. Slide 15 of 14
TOPIC
Example;
Suppose there are 10,000 students at a university.
2,000 are seniors (S), 3,500 are female (F), 800 are seniors & female.
Determine the probability that a randomly selected
student is (1) a senior, (2) female, (3) a senior & female.
1. Pr(S) = 2,000/10,000 = 0.2
2. Pr(F) = 3,500/10,000 = 0.35
3. Pr(S∩F) = 800/10,000 = 0.08
16. Slide 16 of 14
TOPIC
Use the definition of conditional probability
P(A|B) = P(A∩B) / P(B)
& the previously calculated information
P(S) = 0.2; P(F) = 0.35; P(S∩F) = 0.08
to answer the questions below.
1. If a randomly selected student is female, what is the probability that
she is a senior?
P(S|F) = P(S∩F) / P(F)
= 0.08 / 0.35 = 0.228
2. If a randomly selected student is a senior, what is the probability the
student is female?
P (F|S) = P (F∩S) / P (S)
= 0.08 / 0.2 = 0.4
Notice that S∩F = F∩S, so the numerators are the same, but the
denominators are different.
17. Slide 17 of 14
TOPIC
A, B, C are mutually independent
P(A U B) = P(A) + P(B) – P(A) P(B)
P(A U B U C) = P(A) + P(B) +P(C) - P(A) P(B) – P(B) P(C) - P(C) P(A) +
P(A) P(B)P(C)
Formula – Very Important
P(A (B U C)) = P(A) − P(A ∩ B) − P(A ∩ C) + P(A ∩ B ∩ C)
18. Slide 18 of 14
TOPIC
De Morgan’s law
P(Ac ∩ Bc) = P((A U B)c) = 1 − P(A U B)
P(Ac ∩ B) = 1 – P(A U Bc)
19. Slide 19 of 14
TOPIC
PROBLEM
One ball is drawn randomly from a bowl containing four balls numbered
1, 2, 3, and 4. Define the following three events:
Let A be the event that a 1 or 2 is drawn. That is, A = {1, 2}.
Let B be the event that a 1 or 3 is drawn. That is, B = {1, 3}.
Let C be the event that a 1 or 4 is drawn. That is, C = {1, 4}.
Are events A, B, and C pairwise independent? Are they mutually
independent?
