CONDITIONAL PROBABILITY computing mt.pptx

CONDITIONAL PROBABILITY
Suppose E is an event in a sample space S with P(E) >
0. The probability that an event A occurs once E has
occurred (the conditional probability of A given E).
written P(A|E), is defined as
P(A|E) = P(A ∩ E)
P(E)
P(A|E) = number of elements in A ∩ E = n(A ∩ E)
number of elements in E n(E)

EXAMPLE 7.7
A pair of fair dice is tossed. The sample space S consists of the
36 ordered pairs (a, b); thus the probability of any point is 1/36 .
Find the probability that one of the dice is 2 if the sum is 6. That
is, find P(A|E) where E = {sum is 6} and A = {2 appears on at
least one die}
Now E consists of 5 elements and A ∩ E consists of two
elements; namely
E = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} and A ∩ E = {(2, 4), (4, 2)}
P(A|E) = 2/5.
On the other hand A itself consists of 11 elements, that is, A =
{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5,
2), (6, 2)} ; Since S consists of 36 elements, P(A) = 11/36

(b) A couple has two children; the sample space is S
= {bb, bg, gb, gg} with probability ¼ for each point.
Find the probability p that both children are boys if it
is known that:
(i) at least one of the children is a boy;
(ii) the older child is a boy.
(i) Here the reduced space consists of three
elements, {bb, bg, gb}; hence p = 1/3
(ii) Here the reduced space consists of only two
elements {bb, bg}; hence p = 1/2

Multiplication Theorem for Conditional Probability
Suppose A and B are events in a sample space S
with P(A) > 0.
P(B|A) = P(A ∩ B)
P(A)
Multiplying both sides by P(A) gives
P(A ∩ B) = P(A).P(B|A) (Multiplication Theorem for
Conditional Probability)

A lot contains 12 items of which 4 are defective. Three
items are drawn at random from the lot one after the other.
Find the probability p that all three are non defective.
The probability that the first item is non defective is 8/12
since 8 of 12 items are non defective. If the first item is non
defective, then the probability that the next item is non
defective is 7/11 since only 7 of the remaining 11 items are
non defective. If the first 2 items are non defective, then the
probability that the last item is non defective is 6/10 since
only 6 of the remaining 10 items are now non defective.
Thus by the multiplication theorem,
p = 8/12 ・ 7/11 ・ 6/10 = 14/55 ≈ 0.25

INDEPENDENT EVENTS
Events A and B in a probability space S are said to be
independent if the occurrence of one of them does not
influence the occurrence of the other; i.e. B is
independent of A if P(B) is the same as P(B|A).
Substituting P(B) for P(B|A) in the Multiplication
Theorem P(A ∩ B) = P(A).P(B|A) yields P(A ∩ B) =
P(A).P(B).
Definition: Events A and B are independent if
P(A ∩ B) = P(A)P(B); otherwise they are dependent.
Independence is a symmetric relation. In particular, the
equation: P(A ∩ B) = P(A)P(B) implies both P(B|A) =
P(B) and P(A|B) = P(A)

INDEPENDENT EVENTS
EXAMPLE 7.9
A fair coin is tossed three times yielding the equiprobable
space S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT }
Consider the events:
A = {first toss is heads} = {HHH,HHT,HTH,HTT }
B = {second toss is heads) = {HHH,HHT,THH,THT}
C = {exactly two heads in a row} = {HHT,THH}
A and B are independent events. On the other hand, the
relationship between A and C and between B and C is
not obvious. We claim that A and C are independent, but
that B and C are dependent. We have:
P(A) = 4/8 = ½, P(B) = 4/8 = ½ , P(C) = 2/8 = ¼

INDEPENDENT EVENTS
Also,
P(A∩B) = P({HHH,HHT}) = ¼
P(A∩C) = P({HHT}) = 1/8
P(B ∩ C) = P({HHT,THH}) = ¼
Accordingly,
P(A).P(B) = ½ ・ ½ = ¼ = P(A ∩ B), and so A and B are
independent
P(A)P(C) = ½ ・ ¼ = 1/8 = P(A ∩ C), and so A and C
are independent
P(B)P(C) = ½ ・ ¼ = 1/8 ≠ P(B ∩ C), and so B and C
are dependent

INDEPENDENT EVENTS
EXAMPLE 7.10
The probability that A hits a target is ¼ and the probability that
B hits the target is 2/5 . Both shoot at the target. Find the
probability that at least one of them hits the target, i.e., that A
or B (or both) hit the target.
We are given that P(A) = ½ and P(B) = 2/5 , and we seek
P(A∪B).
Furthermore, the probability that A or B hits the target is not
influenced by what the other does; that is, the event that A hits
the target is independent of the event that B hits the target,
that is, P(A ∩ B) = P(A).P(B). Thus
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= P(A) + P(B) − P(A)P (B)
= ¼ + 2/5 − ¼ . 2/5 = 11/20

REVISION QUESTIONS
15. A pair of fair dice is thrown. Find the probability that the sum is 10
or greater if:
(a) 5 appears on the first die;
(b) 5 appears on at least one die.
16. In a certain college town, 25% of the students failed mathematics
(M), 15% failed chemistry (C), and 10% failed both mathematics and
chemistry. A student is selected at random.
a) If he failed chemistry, find the probability that he also failed
mathematics.
b) If he failed mathematics, find the probability that he also failed
chemistry.
c) Find the probability that he failed mathematics or chemistry.
d) Find the probability that he failed neither mathematics nor
chemistry.

17. A pair of fair dice is thrown. Given that the two numbers
appearing are different, find the probability p that:
(a) the sum is 6;
(b) a one appears;
(c) the sum is 4 or less.
18. A class has 12 boys and 4 girls. Suppose three students are
selected at random from the class. Find the probability p that
they are all boys.
19. The probability that A hits a target is 1/3 and the probability
that B hits a target is 1/5 . They both fire at the target. Find the
probability that:
(d) A does not hit the target; (b) both hit the target
(c) one of them hits the target; (d) neither hits the target.

INDEPENDENCE
20. Consider the following events for a family with children:
A = {children of both sexes}, B= {at most one boy}.
(a) Show that A and B are independent events if a family has
three children.
(b) Show that A and B are dependent events if a family has only
two children.
21. Box A contains five red marbles and three blue marbles,
and box B contains three red and two blue.
A marble is drawn at random from each box.
(a) Find the probability p that both marbles are red.
(b) Find the probability p that one is red and one is blue.
22. Prove: If A and B are independent events, then Ac
and Bc
are independent events.

INDEPENDENT REPEATED TRIALS, BINOMIAL DISTRIBUTION
Let S be a finite probability space. The space of n
independent repeated trials means the probability
space Sn consisting of ordered n-tuples of elements of
S, with the probability of an n-tuple defined to be the
product of the probabilities of its components:
P((s1, s2, . . . , sn)) = P(s1).P (s2) . . . P (sn)
EXAMPLE 7.11
Whenever three horses a, b, and c race together, their
respective probabilities of winning are1/2, 1/3, and 1/6
In other words, S = {a, b, c} with P(a) = ½ , P(b) = 1/3,
and P(c) = 1/6

If the horses race twice, then the sample space of the
two repeated trials is S2 = {aa, ab, ac, ba, bb, bc, ca,
cb, cc}.
The probability of each point in S2 is
P(aa) = P(a)P(a) = ½ . ½ = ¼ , P(ba) =1/6, P(ca) = 1/12
P(ab) = P(a)P(b) = ½ .1/3 = 1/6, P(bb) = 1/9, P(cb) = 1/18
P(ac) = P(a)P(c) = ½ .1/6 = 1/12 , P(bc) = 1/18 , P(cc) = 1/36
Thus the probability of c winning the first race and a
winning the second race is P(ca) = 1/12 .

Repeated Trials with Two Outcomes, Bernoulli
Trials, Binomial Experiment
 Assume an experiment with only two outcomes.
Independent repeated trials of such an experiment
are called Bernoulli trials.
 Independent trials means that the outcome of any
trial does not depend on the previous outcomes
(such as tossing a coin). One of the outcomes is
called a success and the other outcome failure.
 If p is the probability of success in a Bernoulli trial,
q = 1 − p is the probability of failure.

• A binomial experiment consists of a fixed number of
Bernoulli trials. A binomial experiment with n trials
and probability p of success is denoted by B(n, p)
• Frequently, we are interested in the number of
successes in a binomial experiment and not in the
order in which they occur.

Theorem 7.7
The probability of exactly k successes in a binomial
experiment B(n, p) is given by
The probability of one or more successes is 1 − qn
.

EXAMPLE 7.12 A fair coin is tossed 6 times; call heads a
success. This is a binomial experiment with n = 6 and p = q = ½
(a) The probability that exactly two heads occurs (i.e., k = 2) is
(b) The probability of getting at least four heads (i.e., k = 4, 5 or
6) is
6 1 4
1 2
6 1 5
1 1
6 1 6
1 0
P(4)+p(5)+p(6) = 4 2 2 + 5 2 2 + 6 2 2
= 15 + 6 + 1 ≈ 0.34
64 64 64

(c) The probability of getting no heads (i.e., all failures) is
so the probability of one or more heads is

REPEATED TRIALS, BINOMIAL DISTRIBUTION
23. Suppose that, whenever horses a, b, c, d race
together, their respective probabilities of winning are
0.2, 0.5, 0.1, 0.2. That is, S= {a, b, c, d} where P(a) =
0.2, P(b) = 0.5, P(c) = 0.1, P(d) = 0.2. They race three
times.
(a) Describe and find the number of elements in the
product probability space S3.
(b) Find the probability that the same horse wins all
three races.
(c) Find the probability that a, b, c each win one race.

For notational convenience, we write xyz for (x, y, z).
(a) By definition, S3 = S × S × S = {xyz | x, y, z ∈ S} and
P(xyz) = P(x)P(y)P(z).
Thus, in particular, S3 contains 43
= 64 elements.
(b) We seek the probability of the event A = {aaa, bbb, ccc,
ddd}. By definition,
P(aaa) = (0.2)3
= 0.008, P(ccc) = (0.1)3
= 0.001
P(bbb) = (0.5)3
= 0.125, P(ddd) = (0.2)3
= 0.008
Thus P(A) = 0.0008 + 0.125 + 0.001 + 0.008 = 0.142.
(c) We seek the probability of the event B = {abc, acb, bac,
bca, cab, cba}. Every element in B has the same
probability, the product (0.2)(0.5)(0.1) = 0.01. Thus P(B) =
6(0.01) = 0.06.

24. The probability that John hits a target is p = ¼. He
fires n = 6 times. Find the probability that he hits the
target:
(a) exactly two times; (b) more than four times; (c) at
least once.
25. A family has six children. Find the probability p that
there are: (a) three boys and three girls; (b) fewer boys
than girls. Assume that the probability of any particular
child being a boy is ½
26. A man fires at a target n = 6 times and hits it k = 2
times, (a) List the different ways that this can happen,
(b) How many ways are there?

7.67. Whenever horses a, b, and c race together, their
respective probabilities of winning are 0.3, 0.5, and
0.2. They race three times.
(a) Find the probability that the same horse wins all
three races.
(b) Find the probability that a, b, c each win one race
7.68. The batting average of a baseball player is
0.300. He comes to bat four times. Find the probability
that he will get:
(a) exactly two hits; (b) at least one hit.

7.69. The probability that Tom scores on a three-point
basketball shot is p = 0.4. He shoots n = 5 times. Find
the probability that he scores: (a) exactly two times; (b)
at least once.
7.70. A certain type of missile hits its target with
probability P = 1/3
(a) If three missiles are fired, find the probability that
the target is hit at least once.
(b) Find the number of missiles that should be fired so
that there is at least a 90% probability of hitting the
target.

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