Virtual Laboratories > Probability Spaces > 1 2 3 4 [5] 6 7 8
As usual, suppose that we have a random experiment
with sample space S, and probability
measure P. Suppose also that we know that an event B in a
has occurred. In general, this information should
clearly alter the probabilities that we assign to other events. In particular, if A is
another event then A occurs if and only if A and B occur;
effectively, the sample space has been reduced to reduced to B. Thus, the
probability of A, given that we know B has occurred, should be proportional
to P(A 
B).
However, conditional probability, given that B has occurred, should still be a
probability measure, that is, it must satisfy the axioms of
probability. This forces the proportionality constant to be 1 / P(B). Thus, we
are led inexorably to the following definition:
Let A and B be events in a random experiment with P(B) > 0. The conditional probability of A given B is defined to be
P(A | B) = P(A B) / P(B).
This argument was based on the axiomatic definition of probability. Let’s explore the idea of conditional probability from the less formal and more intuitive notion of relative frequency. Thus, suppose that we replicate the experiment repeatedly. For an arbitrary event C, let Nn(C) denote the number of times C occurs in the first n runs.
If Nn(B) is large, the conditional probability that A has occurred given that B has occurred should be close to the conditional relative frequency of A given B, namely the relative frequency of A for the runs on which B occurred:
Nn(A B)
/ Nn(B).
But by another application of the relative frequency idea,
P(A 
.
so again we are led to the same definition.
In some cases, conditional probabilities can be computed directly, by effectively reducing the sample space to the given event. In other cases, the formula above is better.
1. Show that as a
function of A, for fixed B, P(A | B) is a probability
measure.
Exercise 1 is the most important property of conditional probability because it means that any result that holds for probability measures in general holds for conditional probability, as long as the conditioning event remains fixed.
2. Suppose that A
and B are events in a random experiment with P(B) > 0.
Prove each of the following:
A then P(A
| B) = 1. B then P(A
| B) = P(A) / P(B). 3. Suppose that A
and B are events in a random experiment, each having positive probability. Show
that
P(B | A) > P(B)
P(A
B) > P(A)P(B)
P(B | A) < P(B)
P(A
B) < P(A)P(B)
P(B | A) = P(B)
P(A
B) = P(A)P(B)In case (a), A and B are said to be positively correlated. Intuitively, the occurrence of either event means that the other event is more likely. In case (b), A and B are said to be negatively correlated. Intuitively, the occurrence of either event means that the other event is less likely. In case (c), A and B are said to be independent. Intuitively, the occurrence of either event does not change the probability of the other event.
Sometimes conditional probabilities are known and can be used to find the probabilities of other events.
4. Suppose that A1,
A2, ..., An are events in a random experiment
whose intersection has positive probability. Prove the multiplication rule of probability.
P(A1 A2 ··· An) = P(A1)P(A2 | A1)P(A3 | A1 A2) ··· P(An | A1 A2 ··· An-1)
The multiplication rule is particularly useful for experiments that consist of dependent stages, where Ai is an event in stage i. Compare the multiplication rule of probability with the multiplication rule of combinatorics.
5.
Suppose that A
and B are events in an experiment with P(A) = 1 / 3, P(B) = 1 / 4, P(A
B) = 1 / 10.
Find each of the following:
6. Consider the experiment
that consists of rolling 2 fair dice and recording the
sequence of scores
Correlation is not transitive. From the previous exercise, for example, note that {X1 = 3}, {Y = 5} are positively correlated, {Y = 5}, {X1 = 2} are positively correlated, but {X1 = 3}, {X1 = 2} are negatively correlated.
7. In dice
experiment, set n = 2. Run the experiment 500 times.
Compute the empirical conditional probabilities corresponding to the conditional
probabilities in the last exercise.
8.
Consider the card experiment that consists of dealing 2 cards from a
standard deck and recording the sequence of cards dealt. For i = 1, 2,
let Qi be the event that card i is a queen and Hi
the event that card i is a heart. For each of the following pairs of
events, compute the probability of each event,
and the conditional probability of each event given the other. Determine whether the events are
positively correlated, negatively correlated, or independent.
9. In the card
experiment, set n = 2. Run the experiment 500 times.
Compute the conditional relative frequencies corresponding to the conditional
probabilities in the last exercise.
10. Consider the
card experiment with n = 3 cards. Find the probability of the following
events:
11. In the
card
experiment, set n = 3 and run the simulation 1000 times.
Compute the empirical probability of each event in the previous exercise and
compare with
the true probability.
12.
In a certain population, 30% of the persons smoke and 8% have a certain type of
heart disease. Moreover, 12% of the persons who smoke have the disease.
13. Suppose that A, B, and C are events in a random experiment
with
B | C) = 1 / 4.
Bc
| C)
B |
C)
Bc | C). 14.
Suppose that A and B are events in a random experiment with P(A)
= 1 / 2, P(B) = 1 /3 , P(A | B) = 3 / 4.
Find each of the following
B).
B).
Ac). 
15. For the M&M data
set, find the empirical probability that a bag
has at least 10 reds, given that the weight of the bag is at least 48 grams.
16. For the Cicada
data,
Again, suppose that we have an experiment with sample space S and probability measure P. Suppose that X is a random variable for the experiment that takes values in a set T. Recall that the probability distribution of X is the probability measure on T given by
P(X
B) for B
T.
Analogously, if A is an event with positive probability, the conditional distribution of X given A is the probability measure on T given by
P(X
B | A)
for B
T.
17.
Consider the experiment
that consists of rolling 2 fair dice and recording the
sequence of scores
18.
Suppose that the time X required to perform a certain job (in minutes) is
uniformly distributed on the interval
19. Recall that Buffon's coin experiment consists of tossing a coin with
radius r 
1/2 randomly on a floor covered
with square tiles of side length 1. The coordinates (X, Y) of the center
of the coin are recorded relative to axes through the center of the square, parallel to
the sides.
20. Run Buffon's coin experiment 500 times. Compute the empirical
probability that
Suppose that {Aj: j
J} is a countable collection of events that that partition the sample space S.
Let B be another event and
suppose that we know P(Aj) and P(B | Aj)
for each j J.
21.
Prove
the the law of total probability:
P(B) = 
j
P(Aj)
P(B | Aj).
22.
Prove
Bayes' Theorem, named after Thomas Bayes:
for k J,
P(Ak | B) = P(Ak)P(B | Ak)
/ j
P(Aj)
P(B | Aj).
In the context of Bayes theorem, P(Aj) is the prior probability of Aj and P(Aj | B) is the posterior distribution of Aj. We will study more general versions of the law of total probability and Bayes theorem in the chapter on Distributions.
23. In the
die-coin experiment, a fair die is rolled and then a fair coin is tossed the number of
times showing on the die.
24. Run the die-coin experiment 200 times.
25. Suppose
that a bag contains 12 coins: 5 are fair, 4 are biased with probability of heads
1/3; and
3 are two-headed. A coin is chosen at random from the bag and tossed.
Compare Exercises 23 and 25. In Exercise 23, we toss a coin with a fixed probability of heads a random number of times. In Exercise 25, we effectively toss a coin with a random probability of heads a fixed number of times.
26. In the
coin-die experiment, a fair coin is tossed. If the coin lands tails, a fair die is rolled.
If the coin lands heads, an ace-six flat die is tossed (1 and 6 have probability 1/4 each,
while 2, 3, 4, 5 have probability 1/8 each).
27. Run the
coin-die experiment
500 times.
28. A plant has 3
assembly lines that produces memory chips. Line 1 produces 50% of the chips and has a
defective rate of 4%; line 2 has produces 30% of the chips and has a defective rate of 5%;
line 3 produces 20% of the chips and has a defective rate of 1%. A chip is chosen at
random from the plant.
29.
The most common form of colorblindness (dichromatism) is a
sex-linked hereditary condition caused by a defect on the X chromosome.
Thus, it is much more common in males than females; 7% of males are colorblind but only 0.5% of females are colorblind. (For more on sex-linked
hereditary disorders, see the discussion of hemophilia.)
In a certain population, 50% are male and 50% are female.
30. An urn
initially contains 6 red and 4 green balls. A ball is chosen at random and
then replaced along with 2 additional balls of the same color; the process
is repeated. This an example of Pólya's urn scheme, named after George
Pólya.
31. Urn 1
contains 4 red and 6 green balls while urn 2 contains 7 red and 3 green balls. An urn is chosen at random and then a
ball is chosen from the
selected urn.
32.
Urn 1 contains 4 red and 6 green balls while urn 2 contains 6 red and 3
green balls. A ball is selected at random from urn 1 and transferred
to urn 2. Then a ball is selected at random from urn 2.
Suppose that we have a random experiment with an event A of interest. When we run the experiment, of course, event A will either occur or not occur. However, we are not able to observe the occurrence or non-occurrence of A directly. Instead we have a test designed to indicate the occurrence of event A; thus the test that can be either positive for A or negative for A. The test also has an element of randomness, and in particular can be in error. Here are some typical examples of the type of situation we have in mind:
Let T be the event that the test is positive for the occurrence of A. The conditional probability P(T | A) is called the sensitivity of the test. The complementary probability
P(Tc | A) = 1 - P(T | A)
is the false negative probability. The conditional probability P(Tc | Ac) is called the specificity of the test. The complementary probability
P(T | Ac) = 1 - P(Tc | Ac)
is the false positive probability. In many cases, the sensitivity and specificity of the test are known, as a result of the development of the test. However, the user of the test is interested in the opposite conditional probabilities:
P(A | T), P(Ac|Tc).
33. Use Bayes'
Theorem to show that
P(A | T) = P(T | A)P(A) / [P(T | A)P(A) + P(T | Ac)P(Ac)].
For a concrete example, suppose that the sensitivity of the test is 0.99 and the specificity of the test is 0.95. Superficially, the test looks good.
34.
Find P(A | T) as a function of p = P(A).
Show that the graph has the following shape:
35.
Show that P(A | T) as a function of P(A)
has the values given in the following table:
| P(A) | P(A | T) |
|---|---|
| 0.001 | 0.019 |
| 0.01 | 0.167 |
| 0.1 | 0.688 |
| 0.2 | 0.832 |
| 0.3 | 0.895 |
The small value of P(A | T) for small values of P(A) is striking. The moral, of course, is that P(A | T) depends critically on P(A), not just on the sensitivity and specificity of the test. Moreover, the correct comparison is P(A | T) with P(A), as in the table, not P(A | T) with P(T | A).