Theorem 8

1
Lecture 2
2

• Theorem 8
• Continuity of Probability Function
• For any increasing or decreasing sequence of
  events
• En, n 1 P(En) P( En)
• Exs
• In a certain town the No. of people with Blood O
  and A are roughly same. The No. of people with B
  is 1/10 of those with A and twice the No. of type
  AB.
• Find Prob. that next baby born in town has type
  AB

3

• Let E 0, F A, G B, H AB
• P(E) P(F) by stmt
• P(G) 1/10 P(F) by stmt these imply P(E)
  P(F) 20 P(H)
• P(G) 2xP(H) by stmt
• Thus, from P(E) P(F) P(G) P(H) 1
• We obtain 20 P(H) 20 P(H) 2 P(H) P(H) 1
• which gives P(H) 1/43

4

• Definition A point is said to be randomly
  selected from interval (a,b) if any 2
  subintervals of (a,b) that have same length are
  equally likely to include the point.
• P(E)
• In the sub-interval (a,ß)

5

• Exs
• Every new book by a certain publisher captures
  randomly between 4-12 of the market. Whats the
  Probability that next book by publisher captures
  at most 6.35 market?

6

• Exs.
• A bookstore gets 6 boxes of books/month on 6
  random days of each month. Suppose 2 boxes are
  from 1 publisher, 2 from another, and 2 left from
  another.
• Define sample space for possible orders in which
  the boxes are received in a given month.
• Describe the event that last 2 boxes from last
  month are from same publisher.
• ? ai box of books, i publisher 1,2,3
• Thus, S x1x2x3x4x5x6 2 of xis are a1, 2 are
  a2, 2 are a3
• And E x1x2x3x4x5x6 S x5 x6

7

• Review
• Counting Principle
• If set E has n elements there are nm ways we can
  pick first from E, then
  from F
• If set F has m elements
• Generalized counting principle
• Let E1. . . Ek be sets with n1. . . nk elements
  respectively
• there are n1n2n3. . . nk ways we can pick from
  E1, then E2, then E3, . . . to Ek (one at a time)
• Theorem A set with n elements has 2n subsets

8

• Permutations
• The number of distinguishable permutations of n
  objects of K diff. types, where n1 are alike, n2,
  n3,
• n n1 n2 n3 . . . nk is

9

• Combinations An unordered arrangement of r
  objects from set A containing n objects (r
  n) is called an r element combination of A
  Given by
• Number of subsets of size r
• that can be built from set size n
  (n choose r)

10

• Review
• Note that
• For any
• Exs
• Floridas Lotto, pick 6 different integers from 1
  to 49 (any order)
• Six Nos are the winning ones. Grand prize for
  all 6/6
• Second prize for all 5/6
• Third prize for all 4/6
• Find probability that a certain choice of bettor
  wins 1st, 2nd, 3rd prizes respectively.

11

• Probablity (1st) 1/13.9M
• Probability (2nd) 258/13.9m
  1/54k
• Probability (3rd) 13.5k/13.9m
  1/1k

12

• Binomial Expansion for any integer n .GE.
  0
• Exs
• Find coeff. of x3y4 in the expression of (2x
  4y)7
• Let u 2x v 4y (2x 4y)7 (u v)7.
• coeff. u3v4 in the expansion of (2x 4y)7 is
  and u3v4 (23 44)
• u3v4 (23 44) 71.6k

13

• Conditional Probability
• If P(B) 0, the conditional probability of A
  given B denoted by P(AB) is
• P(AB)
• if P(B) 0 then zero divide. . .
• The conditional probability of A given that B has
  occurred is the ratio of the probability of joint
  occurrence of A and B and the probability of B

14

• Exs
• We draw 8 cards at random from a 52 card deck.
  Given that 3 of them are spades, whats the
  probability that remaining 5 are also spades?
• Let B be event that at least 3 are spades,
• Let A be event that all of 8 cards are spades,
• Thus, P(AB) is desired result, so
• P(AB) P(AB) / P(B)
• So, P(AB)

15

• But since AB is event that all 8 are spades
• Then, Look in Combinations
• Binomial expansions Stirlings
  formula

16

• Exs
• Suppose 2 fair dice have been tossed and the
  total of their top faces is found to be divisible
  by 5.
• Whats the probability that both of them landed
  at 5?
• Define S (1,4), (2,3), (4,1), (3,2), (4,6),
  (5,5), (6,4)
• Thus P(E5) 1/7

17

• Theorem 1
• Let S be sample space of an experiment, and let B
  be an event of S with P(B) 0, then
• a.) P(AB) 0 for any event A of S
• b.) P(SB) 1
• c.) if A1,A2, . . . is a seq. of mutually
  exclusive events, then
• Clause c.) was used on example about 8 spades.
• Note that sample space can be reduced if,

18

• Note that sample space can be reduced if,
• Let B be event of sample space S with P(B) 0
• A subset A of B Q(A) P(AB), then clearly
  Q(A) 0, Q(B) P(BB) 1 and by theorem 1, if
  A1,A2,A3. . . is a mutually exclusive sequence
• Thus, sample space is reduced from S to B having
  above conditions.

19

• Furthermore, by theorem 1, the function Q, Q(A)
  P(AB) is a probability function thus Q
  satisfies theorems stated for Probability, and
  for all B, P(B) 0, so
• P(FB) 0
• P(ACB) 1 P(AB)
• if C A, then P(ACCB) P(A CB) P(AB)
  P(CB)
• if C A. then P(CB) P(AB)
• P(A U CB) P(AB) P(CB) - P(ACB)
• P(AB) P(ACB) P(ACcB)

20

• To calculate P(A1 A2 A3. . .
  AnB)
• Calculate conditional probability of all possible
  intersections of events in A1,A2, A3, . . . An
  given B
• Add conditional probability obtained by
  intersecting an odd number of events
• Subtract conditional probability obtained by
  intersecting an even number of events.
• For any increasing or decreasing sequence of
  events
• An, n 1

21

• Exs.
• - It is estimated that there are 105 fish in a
  pond of a fish farm 40 are trout, 65 carp.
• Someone got 8 fish
• What is the probability that exactly 2 are trout,
  if we know that at least 3 of them are not?
• A 40, 2
• B 65, 6

22

• A 40, 2
• B 65, 6
• P(B) Probability that exactly 2 are trout,
  excluding the rest

23

• Exs.
• A number is selected at random from 1,2, . . . ,
  10k and is observed to be odd.
• a.) Probability of being divisible by 3
• b.) Probability not divisible by neither 3 nor 5
  ?
• Reduce S 1,3,5,7, . . . 9999, n 5000 odd
  elements
• a.) Since S 5,7,9,11,13, . . . 9999 includes
  exactly 4998/3 1666 odd number divisible by 3
  the reduced sample space has 1667 odd numbers
  divisible by 3, so
• P(3) 1667/5000 0.33

24

• b.)
• Let K be event that No. selected at random is
  odd
• Let F be event that it is divisible by 5
• Let T be event that it is divisible by 3

25

• The desired probability is calculated from
• P(FK) 10,11,12,13, . . . 9995 5000 5
  4995/5
• 999 1 1000
• 1667 1000
  2667/5000 0.53
• 333 x 5000
• x 5000/333 15 (5x3) 5000/15 333

26

• Law of Multiplication Theorem 2
• if P(A1A2A3 . . .An-1) 0 then
• P(A1A2 . . .An-1An) P(A1) P(A2A1) P(A3A1A2).
  . . P(AnA1A2. . .An-1)
• Identities P(AB) P(AB) / P(B) by
  multiplying both sides by P(B) 0
• P(AB) P(B) P(AB)
• This means that probability of joint occurrence
  of A and B is the product of P(B) and P(AB).
• If P(A) 0, then by letting A B and B A
• P(BA) P(A) P(BA) but since
• P(BA) P(AB) this gives P(AB) P(A) P(BA).

27

• Exs.
• There are 5 hills and 6 craters to be explored.
  Mission control assigns them randomly one by one.
• What is the probability that all hills are
  explored first?
• What is probability that hills and craters get
  explored in alternating fashion?
• (C) (H)

28

• Theorem 3 (Law of Total Probability)
• Let B be event with P(B) 0 and P(BC) 0.
  For any event
• A ? P(A) P(AB) P(B) P(ABC) P(BC)
• Useful when not possible to calculate directly
  the probability of occurrences of an event A.
• Theorem 4 (generalization of Theorem 3)
• if B1, B2, . . .Bn is a partition of a sample
  space S and
• P(Bi) 0 for i 1, 2, . . . n then for any
  event A of S
• P(A) P(AB1) P(B1) P(AB2) P(B2) . .
  P(ABn) P(Bn)
• P(ABi) P(Bi)

29

• Caviat make sure B1, B2, . . . Bn form a
  partition (are contained) in sample space S

30

• Theorem 5 (most general form)
• Let S be sample space S. Let B1, B2 . . . be a
  sequence of mutually exclusive events of S such
  that
• If for all i 1, P(Bi) 0, then for any
  event A of S

31

• Identity
• let B be an event of S with P(B) 0
• For a subset A of S, Q(A) P(AB) by theorem 1,
  Q is a probability function.
• If E and F are events of S and P(FB) 0 then
  Q(EF) P(EFB). This can be proven by

32

• Exs
• Suppose that 40 of students on a campus, who are
  married to students of same campus, are female.
  Furthermore, 30 of those students who are
  married, but not to students at this campus, are
  also female. If 1/3 of the married students on
  this campus are married to other students on this
  campus,
• whats the probability that randomly selected
  married student from this campus is female?
• Let M be a random married student
• C random student is married to a student
  same campus
• F random student is female

33

• Let M be a random married student
• C random student is married to a student
  same campus
• F random student is female
• For any event A,
• let Q(A) P(AM) using theorem 1. Q is a
  probability function and applying theorem 3 to Q
  and using last identity we get
• P(FM) P(FMC) P(CM) P(FMCC) P(CCM)
• (0.4)
  (0.30) 0.33