Conditional Probability and Independence

1
Conditional Probability and Independence

• If A and B are events in sample space S and P(B)
  gt 0, then the conditional probability of A given
  B is denoted P(AB) and
• Example. A coin is flipped twice.
  Let S (H,H),(H,T),(T,H),(T,T) and assume all
  four outcomes are equally likely. Let A be the
  event that both flips land on heads and let B be
  the event that at least one flip lands on heads.
  Since A (H,H) and B (H,H), (H,T),
  (T,H), we have

2
Reduced Sample Space

• When working with conditional probability P(AB),
  it is often easier to treat B as the new sample
  space.
• Example. A coin is flipped twice (continued).
  Let S (H,H),(H,T),(T,H),(T,T),
  A (H,H), B (H,H), (H,T),
  (T,H). Now, think
  of B as the sample space, where all outcomes are
  equally likely. Clearly, P(AB) 1/3, which
  agrees with the calculation on the previous
  slide.

3
The Law of Multiplication

• If we multiply through the conditional
  probability of A given B by P(B), we obtain the
  law of multiplication This rule can
  be generalized (see the textbook).
• Problem. Let an urn contain 8 red balls and 4
  white balls. We draw 2 balls from the urn
  without replacement. If we assume that at each
  draw each ball in the urn is equally likely to be
  chosen, what is the probability that both balls
  are red? Solution. Let R1 and R2 denote, resp.,
  the events that the first and second balls are
  red. Using the multiplication rule, we have
  Of course, we could
  solve this problem by a direct count without the
  use of conditional probability.

4
Probability of no king on four draws without
replacement

• Draw from an ordinary deck of 52 cards
• Let Ai be the event that no king is drawn on the
  ith draw.
• This is the same result we previously obtained by
  counting.

5
Law of Total Probability

• Let B be an event with P(B) gt 0 and P(Bc) gt 0.
  Then for any event A,
  This law may also be generalized--see textbook.
• Example. An insurance company rents 35 of the
  cars for its customers from agency I and 65 from
  agency II. If 8 of the cars of agency I and 5
  of the cars of agency II break down during the
  rental periods, what is the probability that a
  car rented by this insurance company breaks down?
  
• A tree diagram is often useful for the law of
  total probability.

6
Bayes Formula--see text for a generalization

• Suppose F1, F2, and F3 are pairwise disjoint and
  SF1?F2 ?F3.
• Now,
• If event E is known to have occurred, then we can
  update the probabilities that the events Fj (the
  hypotheses) will occur by using Bayes formula.
  P(Fj) is called the prior probability of Fj and
  the conditional probability P(Fj E) is the
  posterior probability of Fj after the occurrence
  of E.

7
Example for Bayes formula

• Suppose we have 3 cards which are identical in
  form. The first card has both sides red, the
  second card has both sides black, and the third
  card has one red side and one black side.
  Suppose that one of the cards is randomly
  selected and put down on the ground. If the
  upturned side of the chosen card is red, what is
  the probability that the other side is black?
• Let R2, B2, and M denote the events that the
  chosen card is, resp., all red, all black, and
  mixed (red-black). Letting R be the event that
  the upturned side of the chosen card is red, we
  have

8
Independent events

• In general, P(EF) and P(E) are different. That
  is, knowing that F has occurred generally changes
  the probability of Es occurrence. This leads to
  the following definition.
• Events E and F are independent in case
  If E and F are not independent, we say they
  are dependent.
• Example. Two coins are flipped and all 4
  outcomes are assumed to be equally likely. If E
  is the event that the first coin lands heads and
  F is the event that the second coin lands tails,
  then E and F are independent since

9
More on independent events

• If E and F are independent, so are E and Fc. See
  proof in textbook. What can you say about the
  independence of Ec and Fc?
• Assuming P(EFG) P(E)P(F)P(G) for three events
  E, F, G does not imply pairwise independence.
  See Example 3.29 in the textbook.
• We say E1, E2, , En is independent if for every
  subset
• Example. Suppose we conceive of an experiment
  involving an infinite number of coin flips.
  Suppose Ei is the event that the ith flip turns
  up heads. We believe that these events are
  independent, and this means that all equations of
  type will hold (without the restriction that
  subscripts are n). This shows how to extend
  the concept of independence to a sequence of
  events.

10
ExampleAn experiment with independent
subexperiments

• Independent trials, consisting of rolling a pair
  of dice are performed. What is the probability
  of the event E that we get a sum of 5 before we
  get a sum of 7?
• Let En denote the event that no 5 or 7 appears on
  the first n1 trials and a 5 appears on the nth
  trial. The desired probability is
• Since P(5 on any trial) 4/36 and P(7 on any
  trial) 6/36, by the independence of trials,
  P(En) (1 10/36)n-1(4/36) and thus the desired
  probability is 2/5 using the result of Appendix
  2.
• Let F be event that a 5 occurs on 1st trial, G be
  event that a 7 occurs on 1st trial, and H be
  event that neither 5 nor 7 occurs on 1st trial.
  P(E) P(EF)P(F)P(EG)P(G)P(EH)P(H). Now
  P(EF) 1 and P(EG) 0. P(EH) P(E) since H
  has no effect. We have, P(E) 1/9
  P(E)(13/18) or P(E) 2/5. This is the same
  result as before.

Do you see why ?