L8. Probability Concepts

Question 1

Learning outcomes

Answer 1

1. Define a random variable, an outcome, an event, mutually exclusive events and exhaustive events
2. State the two defining properties of probability and distinguish among empirical, subjective and a priori probabilities
3. State the probability of an event in terms of odds for or against the event
4. Distinguish between unconditional and conditional probabilities
5. Explain the multiplication, addition and total probability rules
6. Calculate and interpret 1) the joint probability of two events 2) the probability that at least one of two events will occur, given the probability of each and the joint probability of the two events 3) a joint probability of any number of independent events
7. Distinguish between dependent and independent events
8. Calculate and interpret an unconditional probability using the total probability rule
9. Explain the use of conditional expectation in investment applications
10. Explain the use of a tree diagram to represent an investment problem
11. Calculate and interpret covariance and correlation and interpret a scatterplot
12. Calculate and interpret the expected value, variance, and standard deviation of a random variable and of a returns on a portfolio
13. Calculate and interpret covariance given a joint probability function
14. Calculate and interpret an updated probability using Bayes’ formula
15. Identify the most appropriate method to solve a particular counting problem and solve counting problems using factorial, combination and permutation concepts

Question 2

Probability, expected value and variance

Answer 2

Random variable - a quantity whose future outcomes are uncertain

Outcomes - a possible value of a random variable

Eg. a portfolio return is 10% a year, manager’s focus may be the likelihood of earning a return that is less than 10%. 10% = particular value or outcome of the random variable portfolio return

Event - specified set of outcomes

in the above example, define the event as the portfolio ears a return below 10% . second event = all possible returns greater than or equal to - 100 percent

Therefore we can define both event as
A = portfolio earns a return of 10%
B = portfolio earns a return below 10%

Question 3

Probability

Answer 3

Probability = a number between 0-1 describing the chance that a stated event will occur

if probability is 0.4 for event B, there is a 40% chance that event will happen

if event impossible, probability = 0
if event certain to happen, probability = 1

2 properties

1. probability of any event E is a number between 0 - 1
2. sum of probabilities of any set of mutually exclusive and exhaustive events equals 1

• mutually exclusive = only 1 event can happen at once
• exhaustive = can cover all possible outcomes

Based on the example above, A & B are mutually exclusive. However if third event introduced C = returns will be more than 10 %, then A, B and C are mutually exclusive and exhaustive events. P(A) + P(B), + P(C) = 1

• probability are often estimated as a relative frequency of occurrence based on historical data = empirical probability grouped as objective probability

Question 4

Other types of probabilities include

Answer 4

1. Subjective probability
   - drawing on subjective or personal judgement (when no reliable historical data available)
2. Priori probability
   - based on logical analysis rather than on observation judgement
   - objective probability similar to empirical probability

Question 5

Interpreting probabilities as odds

Answer 5

eg. 2019 EPS forecast for jetblue ariways ranged from 1.50 to 2.20. Analyst A suggest that the odds for the company beating the highest estimate, $2.20 are 1 to 7. Analyst B argues that odds against that happening are 15 to 1

1. Odds for E = P(E)/[1-P(E)] or probability of E is a/(a+b)
   - the odds for E are the probability of E divided by 1 minus the probability of E

based on above eg, P = 1/(1+7) = 0.125
- this means that for each occurrence of E, we are expecting 7 cases of non-occurrence out of 8 cases

2. Odds against E = [1-P(E)]/P(E) or probability of E is b/(a+b)

based on above eg. P = 1/(15+1) = 0.0625

• if a wager of $1 placed on E, if successful, returns $15 in profits plus the $1 staked in wager
• WIN: Probability = 1/16; profit = $15
• LOSE: Probability = 15/16; profit = -$1

Pairs arbitrage trade = trade in 2 closely related stocks involving the short sale of one and the purchase of another

Dutch book theorem = a result in probability theory stating that inconsistent probabilities create profit opportunities

Question 6

Practice Qns for Odds

Answer 6

If probability of event Z is 14%, the odds for Z is?

Solution
Since P(E)/[1-P(E)] 

So odds = 0.14/(1-0.14) = 0.14/0.86 = 0.163

Question 7

Conditional and joint probabilty

Answer 7

Unconditional probability - probability of an event not conditioned on another event. eg: what is the profitability of this event A? Also known as marginal probability
EG. if there are 1000 portfolios, and 600 of the portfolio return above risk free rate, what is the probability of the event that portfolio return is > that risk free rate?
Numerator = # of observation where Rp > Rf = 60
Denominator = # of observation of Rp = 1000
So probability is 600/1000 = 0.6

Conditional probability = probability of an event (conditioned on) another event. eg: what is the probability of A given that B as occured?
EG: if there are 1000 portfolios, what is the probability of investment > risk free rate GIVEN THAT all return is >0%. Meaning that all portfolio is already positive
Numerator = # of observation of Rp > Rf
Denominator = # of observation at Rp > 0
ie. P(E/B) probability of event given B; so off all the Rp>0, how many is more than the risk free rate?

Joint probability = the probability of the joint occurrence of stated events. eg: what is the probability of A and B happening?

Exact definition of conditional probability = The conditional probability of A given that B has occured is equal to the join probability of A and B divided by the probability of B

P(A|B) = P(AB)/P(B), P(B) ≠ 0
in words this formula means probability of A GIVEN THAT (|) B has happened

Multiplication rule of probability
P(AB) = P(A|B)P(B)

Question 8

Practice Qns for probability

Answer 8

Year 2 Winner Year 2 Loser
Year 1 Winner 65.5% 34.5%
Year 1 Loser 15.5% 84.5%

1. What are the 4 events to define the four conditional probabilities?
2. State the 4 entries of the table as conditional probabilities using P(this event | that event) = number
3. Are the conditional probabilities empirical, priori or subjective?
4. Calculate the probability of the event a fund is a loser in both year 1 and 2

Solution
1. Fund is a year 1 winner, Fund is a year 1 loser, Fund is a year 2 winner, Fund is a year 2 loser

2. P(fund is a year 2 winner | fund is a year 1 winner) = 0.655
   P(fund is a year 2 loser | fund is a year 1 winner) = 0.345
   P(fund is a year 2 winner | fund is a year 1 loser) = 0.155
   P(find is a year 2 loser | fund is a year 1 loser) = 0.0845
3. Empirical as it is calculated based on past data
4. Let A represent event that year 2 loser and B represent the event that fund is a year 1 loser. Therefore event AB is the event that a fund is a loser in both Y1 and Y2.

P(A|B) = 0.845 and P(B) = 0.5 because 50% is loser
Thus, P(AB) = P(A|B)P(B)
= 0.845(0.5) = 0.4225 or 0.423

Question 9

Additional rule for probabilities

Answer 9

Principle that states that the probability that A or B occurs equals the probability that A occurs, plus the probability that B occurs, minus the probability that both A and B occur

P(A or B) = P(A) + P(B) - P(AB)

Question 10

Practice Qns for probability

Answer 10

Order 1 was placed at a price limit of $10 and the probability that it will execute within an hour is 0.35. Order 2 was placed at a price limit of $9.75 and has 0.25 probability of executing within the same hour

1. What is the probability that either order 1 or 2 will execute?

Solution
We want to find P(AorB) = P(A) + P(B) - P(AB) 
We know that P(A) is 0.35, P(B) is 0.25
P(AB) = P(A|B)(PB) 
P(AB) = 1*0.25 = 0.25
~ why 1 because if order 2 executes, order 1 sure execute because price must pass through 10 to reach 9.75
Therefore, 0.35+0.25 - 0.25 = 0.35 
P(AorB) = 0.35

2. What is the probability that order 2 executes given that order 1 executes?
   Solution
   We want to find P(B|A) = P(BA) / P(A)
   P(B|A) = 0.25 / 0.35 = 0.714

Question 11

Dependent and independent events

Answer 11

Independent events - the occurrence of one event does not affect the probability of another event occurring
P(A|B) = P(A) or P(B|A) = P(B)
Multiplication rule = P(AB) = P(A)P(B)
eg if 2 events are independent with probabilities of 0.75 and 0.50 respectively, then the probability of both occurring is 0.375 = 0.75(0.5)

Dependent events = the occurrence of one event depends (is related to) the occurrence of another

Total probability rule
P(A) = P(A|S)P(S) + P(A|S^c)P(S^c)

Question 12

Practice Qns for probability

Answer 12

Suppose that 5 percent of the stocks meeting your stock selection criteria are in telecom industry. Also dividend paying stocks are 1% of total number of stocks meeting your selection criteria. What is the probability that a stock is dividend paying, given that it is a stock that met your criteria?

= 0.01/0.05 = 0.2

Question 13

Practice Qns for probability

Answer 13

You are using the 3 criteria to screen potential acquisitions from a list of 500 companies.

C1 - product line compatible = 0.2
C2 - Company will increase combined sales rate = 0.45
C3 - balance sheet impact manageable = 0.78

If criteria are independent, how many companies pass the screen?

P(ABC) = 0.2 x 0.45 x 0.78 = 0.0702
0.0702 x 500 = 35.10 = 35

Question 14

Practice Qns for probability

Answer 14

Apply both valuation criteria and financial strength criteria when choosing stocks. Probability of randomly selected stock meets your valuation criteria = 0.25. Given that a stock meets your valuation criteria, probability that stock meets your financial strength criteria is 0.4. What is the probability that stock meets both your valuation and financial strength criteria?

P(AB) = P(A|B)P(B) given that A = financial strength and B = valuation

So it is 0.4 * 0.25 = 0.10

Question 15

Practice Qns for probability

Answer 15

4 valuation screen and there are 1200 investments. Expected number to pass all 4 screens is?

Screen 1 = 0.65 (probability of passing)
Screen 2 = 0.45
Screen 3 = 0.40
Screen 4 = 0.3

P(ABCD) = 0.65 x 0.45 x 0.4 x 0.3 = 0.0351
0.0351 * 1200 = 42.12 = 42 investments

Question 16

Practice Qns for probability

Answer 16

Analysts develop criteria to evaluate distressed credits

40% of companies tested will go bankrupt within a year: P(non survivor) = 0.4
55% of companies tested will pass: P(pass test) = 0.55
85% probability that a company will pass the test given that is survives a year: P(pass test|survivor) = 0.85

Using total probability rule, probability that a company passed the test given that it goes bankrupt can be determined. P(pass test| non survivor) = ?

Solution
P(A) = P(A|S)P(S) + P(A|S^c)P(S^c)
Given that P(non survivor) = 0.4 then P(survivor) = 0.6

P(pass test) = P(pass test | survivor) + P(pass test | non survivor)P(non survivor) 
0.55 = 0.85(0.6) + P(pass test | non survivor)(0.4) 
Thus P(pass test | non survivor) = [0.55-0.85(0.6)]/0.4 = 0.1

Question 17

Expected value (mean)

Answer 17

Expected value: expected value of a random variable is the probability weighted average of the possible outcomes of the random variable. For a random variable X, the expected value of X is denoted E(X)

• looks either to the future as a forecast or the true value of the mean

EG.
Bankcorp’s EPA for current fiscal year
Probability vs EPS ($)

0. 15, $2.60
1. 45, $2.45
2. 24, $2.20
3. 16, $2

Total probability = 1

What is the expected value of EPS for current fiscal year?

E(EPS) = 0.15(2.60) + 0.45(2.45) + 0.24(2.20) + 0.16(2)
= $2.3405

Equation
E(X) = P(X1)X1 + P(X2)X2….+P(Xn)Xn

Question 18

Variance and ST for expected value

Answer 18

Expected value is forecast, we cannot count of an individual forecast being realised.

• need to measure the risk we face
• variance and SD measure the dispersion of outcomes around the expected value or forecast

Variance = variance of random variable is the expected value (probability weighted average) of squared deviations from the random variable’s expected value

• variance (X) = P{[X1- E(X)]^2} + P{[X2- E(X)]^2} …… + P{[Xn- E(X)]^2}
• Variance is a number greater or = to 0
• if 0 = no dispersion or risk
• increasing variance indicates increasing dispersion

Standard deviation is the positive square root of variance

Eg.
EG.
Bankcorp’s EPA for current fiscal year
Probability vs EPS ($)

0. 15, $2.60
1. 45, $2.45
2. 24, $2.20
3. 16, $2

Total probability = 1

What are the variance and SD?

Variance (EPS) = 0.15(2.6-2.34)^2 + 0.45(2.45-2.34)^2 + 0.24(2.2-2.34)^2 + 0.16(2-2.34)^2
= 0.01014 + 0.005445+0.0047040 + 0.0184960 = 0.038785

SD = √V
= 0.1969391 = 0.2

Question 19

Conditional measures of expected value and variance

Answer 19

Making adjustments of forecast based on new information and events = conditional expected values

Expected value of random variable X given an event S is denoted as E(X | S)

E(X | S) = P(x1 |S)x1 + P(x2 |S)x2+ …. + P(xn |S)xn

Total probability rule for expected value
E(X) = E(X | S1)P(S1) + E(X | S2)P(S2) + .. E(X | Sn)P(Sn)

Question 20

Practice Qns for expected value

Answer 20

E(EPS) = 2.34
Prob of declining interest rate = 0.6
EPS will be $2.60 at Probability = 0.25
Unconditional probability = 0.6 x 0.25 = 0.15
EPS will be $2.45 at Probability = 0.75
Unconditional probability = 0.6 x 0.75 = 0.45

Prob of stable interest rate = 0.4
EPS will be $2.20 at Probability = 0.6
Unconditional probability = 0.4 x 0.6 = 0.24
EPS will be $2.00 at Probability = 0.4
Unconditional probability = 0.4 x 0.4 = 0.16

What is E(EPS | declining interest rate)?
= 0.25(2.6) + 0.75(2.45) = 2.4875

What is E(EPS | stable interest rate)?
= 0.6(2.20) + 0.4(2) = 2.12

Total probability rule for expected value
= E(EPS)
= E(EPS | decline IR)P(deline IR) + E(EPS | stable IR)P(stable IR)
= 2.4875(0.6) + 2.12(0.4) = 2.3405

Variance?
V(EPS | decline IR) = 0.25(2.6-2.4875)^2 + 0.75(2.45-2.4875)^2 = 0.0031641 + 0.0010547 = 0.004219

V(EPS | stable IR) = 0.6(2.2-2.12)^2 + 0.4(2-2.12)^2 = 0.0038400 + 0.0057600 = 0.0096

Question 21

Practice Qns finding expected value using calculator

Answer 21

E(EPS) = 2.34
Prob of declining interest rate = 0.6
EPS will be $2.60 at Probability = 0.25
EPS will be $2.45 at Probability = 0.75

Prob of stable interest rate = 0.4
EPS will be $2.20 at Probability = 0.6
EPS will be $2.00 at Probability = 0.4

What is the expected value for decline IR?
Type 2nd 7
Key in X 1 as 2.6, Y1 as 25% press down
Key in X 2 as 2.45, Y1 as 75% press down
Type 2nd 8 - change to I-V by pressing 2nd Set, 2nd Set till you reach 1-V
Then press down
x = expected value = 2.4875
sigma X = SD = 0.0649519
to find variance ^2 the SD = 0.0042187

What is the expected value for stable IR?
Repeat for stable IR

Question 22

Covariance, correlation and expected value, variance and standard deviation of portfolio returns

Answer 22

Properties of expected value
E(WR) = wE(R)
- expected value of a constant x a random variable

Expected value of a weighted sum of random variables
E(W1R1 + W2R2) = wE1(R1) + wE2(R2) + ….. + wEn(Rn)

Calculation of portfolio expected return
E(Rp) = E(W1R1 + W2R2+ WnRn) = wE1(R1) + wE2(R2) + ….. + wEn(Rn)

Eg. 3 asset class with expected returns
Asset class Weight Expected return(%)
1 0.50 13
2 0.25 6
3 0.25 15

Portfolio expected return
E(Rp) = 0.5 (13) + 0.25(6) + 0.25(15)
E(Rp) = 6.5 + 1.5 + 3.75 = 11.75%

Question 23

Definition of covariance

Answer 23

Population covariance (forward looking) 
Cov(Ri, Rj) = E(Ri - ERi)(Rj - ERj) 

Sample covariance (historical data) 
Cov(Ri, Rj) = sum of(Ri - ERi)(Rj - ERj) /n-1 

• covariance of returns is negative if, return on one asset is above its expected value, the return on the other asset tends to be below its expected value
• covariance of returns is 0 if return on assets are unrelated
• covariance of return is positive when returns on both assets tend to be one the same side (above or below) their expected values at the same time

Question 24

Definition of correlation

Answer 24

Properties of correlation
- range from -1 and + 1 for 2 random variables

• correlation of 0 = absence of any linear relationship between variables
• positive correlation = strong positive linear relationships; one increase, another increase
• negative correlation = strong negative linear relationships; one increase, another decrease

Spurious correlation (correlation that misleadingly points toward associations between variables)

• correlation between 2 variables that reflects chance relationships in a particular data set
• correlation induced by a calculation that mixes each of two variables with a third
• correlation between two variables arising not from a direct relation between them but from the third variable

Correlation is;
P(RiRj) = Cov(Ri,Rj) / SD(Ri,Rj)

Question 25

Practice Qns for portfolio expected return and variance of return

Answer 25

2 mutual funds, 75% invested in A Fund A B E(Ra) = 20% E(Rb) = 12% Covariance matrix Fund A B A 625 120 B 120 196 1. What is the expected return of the portfolio? 0. 75(20) + 0.25(12) = 18% 2. Calculate the correlation matrix for this problem ``` Fund A B A 625(1) 120(3) B 120(4) 196(2) ``` ``` Solution: P(RiRj) = Cov(Ri,Rj) / SD(Ri,Rj) For (1) = 625/25(sqt root of 625 = SD)*25 = 1 For (2) = 196/14*14 = 1 For (3) = 120/25*14) = 0.34 ``` So correlation matrix is; Fund A B A 1 0.34 B 0.34 1 2 variances, and 1 covariance Portfolio return = 0.75^2(625) + 0.25^2(196) + 2 (0.25)(0.75)120 = 351.562 + 12.250 + 45 = 408.812 3. Compute portfolio SD of return SD = √408.812 = 20.22% * NOTATION - sigma = SD - sigma^2 = variance - sigma with subscript = covariance

Question 26

Covariance matrix

Answer 26

Asset A. B C A Var Cov(1,2) Cov(1,3) B Cov(2,1) Var Cov(2,3) C Cov(3,1) Cov(3,2) Var To find matrix, need to find 3 variances, and 3 covariances (since 9-3 = 6 but since its mirrored, so 6/2 = 3) if n assets, then there is n x n = n^2 entries n variation, but (n^2 - n) / 2 covariances

Question 27

Practice qns for covariance matrix

Answer 27

Eg. 3 asset class with expected returns Asset class Weight Expected return(%) 1 0.50 13 2 0.25 6 3 0.25 15 Matrix S&P 500 Bonds MSCI S&P 500 400 45 189 Bonds 45 81 38 MSCI 189 38 441 Portfolio returns = finding 3 variance, 3 covariance = 0.5^2(400) + 0.25^2(81) + 0.25^(441) + 2(0.5)(0.25)45 + 2(0.5)(0.25)189 + 2(0.25)(0.25)38 = 195.875 SD = √196.875 = 14%

Question 28

Covariance given a joint probability function

Answer 28

Joint probability function = function giving the probability of joint occurrences of values of stated random variables Multiplication rule for expected value of the product of uncorrelated random variables ``` P(X,Y) = probability of X happening along with Y P(X|Y) = P(X) given that Y has already happened ```

Question 29

Practice Qns for joint probability function

Answer 29

``` Rb = 20% Rb = 16% Rb = 10% Ra = 25% 0.2 0 0 Ra = 12% 0 0.5 0 Ra = 10% 0 0 0.3 ``` How to read this? What is the likelihood of fund A = 25% and fund B = 20% expected returns? 20% ``` E(Ra) = 0.25(0.2) +0.12(0.5) + 0.10(0.3) = 14% E(Rb) = 0.2(0.2) + 0.16(0.5)+ 0.1(0.3) = 15% ``` Covariance Ra-E(Ra) Rb-E(Rb) Cross product P(Ra,Rb) G 25-14 =11 20-15=5 11 x 5 = 55 0.2 A 12-14 =-2 16-15=1 -2 x 1 = -2 0.5 P 10-14 =-4 10-15=-5 -4 x -5 = 20. 0.3 Joint probability function Cov(A,B) = sum of P(Ra,Rb)(Ra-E(Ra)(Rb-E(Rb) Cov(A,B) = 0.2(55)+0.5(-2)+0.3(20) = 11 -1 +6 = 16

Question 30

Bayes Formula (need to revise again)

Answer 30

Bayes formula (aka inverse probability) = given a set of prior probabilities for an event of interest, if you receive new information, the rule for updating your probability of the event is = (probability of new information given event/ unconditional probability of the new information) x prior probability of event IE P(Event | Information) = [P(Information | event) / P (information)] x P(Event) ``` Eg. estimating EPS These are prior probabilities (prior to arrival of new info) P(EPS exceeded consensus) = 0.45 P(EPS met consensus) = 0.3 P(EPS fell short of consensus) = 0.25 ``` New info = coy expanding = increase sales demand; so now more likely EPS will exceed consensus So now, in light of new info, what is the updated probability that EPS will exceed consensus estimate? So now updated probability as follows: P(DriveMed expands | EPS exceed consensus) = 0.75 P(DriveMed expands | EPS met consensus) = 0.20 P(DriveMed expands | EPS fell short consensus) = 0.05 So these are now conditional probabilities aka likelihoods So combine these conditional probabilities with prior probabilities to get unconditional probability for DriveMed expanding P(DriveMed expands) = P(DriveMed expands | EPS exceed consensus) x P(EPS exceeded consensus) + P(DriveMed expands | EPS met consensus) x P(EPS met consensus) + P(DriveMed expands | EPS fell short consensus) + P(EPS fell short of consensus) = 0.75(0.45) + 0.2(0.3) + 0.05(0.25) = 41% so applying Bayes' formula P(EPS Exceeded consensus | DriveMed expands) = [P(DriveMed expands | EPS exceed consensus) / P(DriveMed expands)] * P(EPS exceeded consensus) = (0.75/0.41)*0.45 = 0.823171 This means that prior to drivemed's announcement, probability that drivemed would beat consensus expectations was 45%. After announcement, probability = 82.3% The updated probability = posterior probability

Question 31

Multiplication rule of counting

Answer 31

Suppose we have 3 steps in an investment decision process. The first step can be done in 2 ways, the second in 4 ways, and the third in 3 ways, following the multiplication rule. there are (2)(4)(3) = 24 ways in which we can carry out the three steps Compact notation for the multiplication we performed is 3! (read as 3 factorial) if we have n analysts, the number of ways we could assign them to n tasks would be n! = n(n-1)(n-2)(n-3)....1

Question 32

Labeling problems

Answer 32

Eg. a mutual fund guide ranked 18 bond mutuals funds by total returns last year. guide also assigned each fund one of 5 risk labels: ``` high risks (4 funds) above-average risks(4 funds) average risk (3 funds) below average risks (4 funds) low risks (3 risks) TOTAL = 18 funds ``` but how many ways can we take the 18 funds and label them as the above guide? there are 18! possible sequences in total there are (4!)(4!)(3!)(4!)(3!) equivalent sequences so we have 18!/[(4!)(4!)(3!)(4!)(3!)] = 12,864,852,000

Question 33

Multinominal formula Combination formula

Answer 33

n!/(n1!n2!....nk!) - the number of ways n objects can be labelled with k different labels nCr = n!/(n-r)!r! - provides the number of ways that r objects can be chosen from a total of n objects, when the order in which the r objects are listed does not matter Eg. 2 sequences of 5 moves containing 3 up moves, such as UUUDD or UDUUD result in same stock price, how many sequences of 5 moves belong to the group with 3 up moves? nCR = 5!/[(5-3)!3!] = 120/12 = 10 ways ``` in calculator to find 5P3 press 5 2nd + (ncr) 3 = ```

Question 34

Permutation formula

Answer 34

nPr = n! / (n-r)! - used to choose r objects from a total of n objects when order matters Suppose jurors want to select 3 companies out of 5 to receive 1st 3 awards, how many ways can jurors make the 3 awards? nPr = 5!/(5-3)! = 120/2 = 60 ways ``` in calculator to find 5P3 press 5 2nd - (npr) 3 = ```

Question 35

Practice Qns for Bayes Formula

Answer 35

Criteria 1. 40% of companies to which the test is administered will go bankrupt within 12 months: P(non survivor) = 0.4 2. 55% companies to which test is administered pass it: P(pass test) = 0.55 3. Probability that a company will pass the test is given that it will subsequently survive 12 months, is 0.85: P(Pass test | survivor) = 0.85 QNS: So what is P(pass test | non survivor)? Solution Total probability rule P(pass test) = P(pass test | survivor) P(survivor) + P(pass test | non survivor ) P(non survivor) ``` Since P(survivor) = 1-0.4 = 0.6 Therefore, P(pass test) = 0.55 = 0.85(0.6) + P(pass test | non survivor)(0.4) ``` Thus P(pass test | non survivor) = [0.55-0.85(0.6)]/0.4 = 0.1 QNS: Using Bayes formula, calculate the probability that a company is a survivor, given that it passes the test; P(survivor | pass test) ``` Solution P(survivor | pass test) = [P(pass test | survivor) / P(pass test)] P (survivor) = (0.85/0.55)0.6 = 0.927 ```