(a) Conditional Probability: Consider the data of Illustration 14-3, Suppose the randomly selected estate surveyor is a female, what is the probability that such a person invested in the petroleum sector? This is a conditional probability which is the probability that a selected person has invested in the petroleum sector given (when it is already known) that the person is a female. This is written as P(selected person has invested in the petroleum sector/ the person is a female).
The symbol / is read as given. By definition, the probability that an event B occurs when it is known that A has already occurred is:
P(B/A) which is defined as:
P(B/A) =
) A ( P
) B A (
P
In other words, the conditional probability that B occurs when it is known that A has previously occurred is the quotient of the probability that A and B occur simultaneously and the unconditional probability that A occurs.
Thus, in the last problem, the probability that the person has invested in the petroleum sector given that the person is a female.
P (Selected Person has invested in the petroleum sector AND the person is also a female)
P(Selected Person is a Female)
35 3 60 / 35
60 / 3
Notice that the conditional probability is 3/35 while the unconditional probability that the selected person has invested in the petroleum sector is 10/60, showing that in general P(B/A)
≠ P(B)
(b) Independence
Suppose P(B/A) = P(B). In other words, the conditional probability of B given that A has previously occurred is the same as the probability that B occurs whether or not A has previously occurred. When this happens, it is clear that A and B are occurring independently. Further, when P(B/A) = P(B), it follows from the definition of conditional probability that P(B/A)=
) (
) (
A P
B A
P --- 14.5.1 When the two events occur independently
Hence, P(A B) = P(A) P(B) --- 14.5.2 Also then P(B/A) = P(B) --- 14.5.3 This is the general definition of independence.
Definition: Events A and B are said to be independent if:
P(A B) = P(A) P(B), otherwise
A and B are dependent or not independent.
Further, since P(B/A) =
) (
) (
A P
B A
P , then
P(A B) = P(A) P(B/A) --- 14.5.4 The expression 14.5.2 holds when A and B are independent and
14.5.4 holds when A and B are dependent. Expression 14.5.4 is called the multiplication law of probability.
(c) Multiplication Law
P(A B)= P(A) P(B/A) --- 14.5.5 for dependent events A and B
But when A and B are independent, then the law simplifies to P(A B) = P(A) P(B)
as obtained in 14.5.2 above.
329 ILLUSTRATION 14-4
There are 7 companies manufacturing table salt (S) and 8 different companies manufacturing beverages (B).
Two companies are to be selected one after the other as training centers for food vendors. Compute the probability that the two randomly selected companies are:
(i) Salt manufacturers
(ii) Manufacturers of the same products
(iii) Manufacturers of different products, when:
(a) The selection process is without replacement (b) The selection is with replacement
SUGGESTED SOLUTION 14-4
(a) Selection is without replacement
(i) Required: the probability that both companies selected are salt manufacturers.
= P(1st company selected is a manufacturer of salt) x P (2nd company selected is a manufacturer of salt/1st chosen is a manufacturer of salt)
P(SS) = P(S) P(S/S), where
P(S) = P (1st chosen company is a manufacturer of salt) P(S/S)= P(2nd selected company is a manufacturer of salt
given that the 1st selected is a manufacturer salt) P(S) = 7/15
If a salt manufacturing company is selected first,
then we have only 6 salt manufacturing companies left for the 2nd choice. Hence:
P(S/S) =
14 6 8 6
6
P(SS) = P(S) P(S/S) =
14 . 6 15
7 = 0.2
(ii) Required: The probability that the two selected companies manufacture the same product
= P(SS) + P(BB)
= P(S) P(S/S) + P(B) P(B/B)
= 14
. 7 15
8 14 . 6 15
7 = 0.47
based on similar argument to the one used in (3)above.
(iii) Required: P(The two selected companies are manufacturers of different products)
= P(SB) + P(BS)
= P(S) P(B/S) + P(B) P(S/B)
=
14 . 7 15
8 14 . 8 15
7 = 0.53
(b) Selection is with replacement
(i) P(Both Selected companies are salt manufacturers)
= P(SS) =P(S)P(S/S). However when selection is with replacement
P(S/S) = 15
7 which is the same as P(S) Thus with replacement P(S/S) = P(S);
P(S/S) = P(S); P(B/S) = P(B) and P(B/B) = P(B) Hence: (i) P(SS) = P(S) P(S) =
2
15
7 = 0.22 (ii) P(SS or BB)= P(SS) + P(BB)
= P(S) P(S) + P(B). P(B)
= 15
8 15
8 15
7 15
7 = 0.5
(iii) P(SB or BS)= P(SB) + P(BS)
= P(S). P(B) + P(B) P(S)
= 15
7 15
8 15
8 15
7 = 0.498
331 (d) Extended Probability law
Let A be an event on a sample space S. Suppose B1, B2 and B3 are a set of mutually exclusive events on S such that at least one of them must occur, then
P(A) = P(B1) P(A/B1) + P(B2) P(A/B2) + P(B3) P(A/B3) We illustrate with the following example
ILLUSTRATION 14-5
An industrialist operates in a community where a year‟s business environmental condition can be classified as favorable (F), Normal (N) or unfavorable (U). Over the years, favorable business environmental condition has prevailed for 40% of the time, while the proportion of years with normal business environmental condition is 35%. The probability that the industrialist will expand the scope of his business in a year with favorable environmental condition is 15%,
while the probabilities of expansion in a year with normal or unfavorable environmental condition are 10% and 2%
respectively. What is the probability that the industrialist will expand the scope of the business in year 2006?
SUGGESTED SOLUTION 14-5 Required: P(Expand business)
= P(Favorable condition) P(Expansion/Favorable condition) +P(Normal condition) P(Expansion/Normal condition)
+P (Unfavorable condition) P(Expansion/Unfavorable condition) From the given information in the question,
P (favorable condition) = 40% = 0.4 P (expansion / favorable condition) = 0.15 P (Normal condition) = 35% = 0.35
P (expansion / Normal condition)= 10% = 0.1 P (unfavorable condition) = 1-0.4-0.35 = 0.25 P (Expansion / unfavorable condition) = 2% = 0.02 Substituting these in the expression above we get
P (expansion in 2006) = (0.40) (0.15) + (.35) (.10) + (.25) (.02)
= 0.06 + 0.035 + 0.005 = 0.1
The data provided for the solution of the problem can be presented as a probability tree as shown in the following diagram.
E Expand P(E|F)
0.15
F P(EC/F) EC 0.95
P(F)
0.40 P(E|N) E P(N) N 0.10
0.35
0.90
P(EC|N) EC P(UF)
UF
P(E|UF) 0.02
0.98 P(EC | UF)
EC Key
E Expand
EC No expansion Fig 14.5.1: A probability tree
Note
(i) For example P(Ec / F) = P(No expansion given favorable condition)
= 1- P(expansion given favorable condition). In other words P(Ec/F) = 1-P(E/F).
(ii) Following our last calculation
P(E) = P(F) P(E/F) + P(N) P(E/N) + P(UF) P(E/UF)
= (0.40) (0.15) + (0.35) (0.10) + 0.25 (0.02) = 0.06 + 0.035 + 0.005 = 0.1
Industrialist 0.25
E
333
which is the sum of the product of probabilities leading to E.
Note that to get to E one can go via F or N or UF
Similarly the probability that the industrialist will not expand is the sum of the products of probabilities leading to Ec (No expansion).
Thus P(Ec) = (0.40) (0.95) + (0.35)(0.90) + (.25)(0.98) = 0.380 + 0.315 + 0.245 = 0.940
(F) Further treatment of independence
Note that two events A and B are independent if P(A B) = P(A) P(B).
This definition is extendable to 3 or more events. For example 3 events A,B and C are independent if P(A and B and C ) = P (A B C) = P(A).P(B).P(C). The following is a practical example of the application of independence.
ILLUSTRATION 14-6
PANFOOD (Nig) LTD is a food processing company located in an area with irregular PHCN power supply.
A day when the company has less than 8 hours power supply is classified as an unproductive day. Past records show that the proportion of unproductive days is 40%.
Assuming that day-by-day power supplies are independent, compute the probability that from Monday to Wednesday of a given week the company will have unproductive days on
(i) Monday only
(ii) Tuesday and Wednesday only (iii) one day only
(iv) two days only (v) all the 3 days (vi) none of the 3 days
(vii) at least one of the 3 days.
SUGGESTED SOLUTION 14-6
Let A denote the event that the company has an unproductive day.
From the data provided:
P(A) = 0.40: Hence P(Ac) = 1 – P(A) = 1-0.4 = 0.6 Thus probability of an unproductive day = 0.4
And probability of a productive day = 0.6
(i) Probability that only Monday is unproductive.
The situation is as shown below:
Mon Tue Wed
A Ac Ac
This means that Mon is unproductive, while Tuesday is productive and Wednesday is productive.
Thus P(Mon only unproductive)
= P (Mon is unproductive AND Tue productive AND Wed is productive)
= P(Mon unproductive)P(Tue productive) P (Wed productive)
= P(A) P(Ac) P(Ac) by independence
= (0.4) (0.6) (0.6) = 0.144
(ii) P (only Tuesday and Wednesday are unproductive)
= P (Monday productive and Tue unproductive and Wed unproductive)
= P (Mon productive) P(The unproductive) P (Wed unproductive)
= P(Ac) P(A) P(A)
= (0.6) (0.4) (0.4) = 0.096
(iii) For the company to be unproductive on only one day there will be:
Mon Tue Wed corresponding probabilities
UP P P (0.4) (0.6) (0.6) = 0.144
P UP P (0.6) (0.4) (0.6) = 0.144
P P UP (0.6) (0.6) (0.4) = 0.144
Thus the probability that the company will be unproductive for only one day
=3(0.4) (0.6)2 = 0.432
335
(iv) For the company to be unproductive for only 2 of the 3 days
Mon Tue Wed PROBABILITY
P UP UP (0.6) (0.4) (0.4) = 0.096
UP P UP (0.4) (0.6) (0.4) = 0.096
UP UP P (0.4) (0.4) (0.6) = 0.096
P(company is unproductive for only two days)
= 3(0.4)2(0.6) = 0.288
(v) Required: P(Unproductive on all of the 3 days)
MON TUE WED
UP UP UP
With probability (0.4) (0.4) (0.4) = (0.4)3 = 0.064 Hence P(Company is unproductive for all the 3 days is)
= (0.4)3= 0.064
(vi) Required: unproductive on none of the three days
Mon Tue Wed
P P P
With probability (0.6) (0.6) (0.6) = 0.216
Hence P (Company is unproductive in none of the 3 days or productive in all the 3days) = (0.6)3 = 0.216
(vii) Note that complement of „NONE‟ is at least one, Hence the probability that the company will be unproductive for at least one of the 3 days
= 1-P(unproductive in none of the three days)