Problem 1:
Bayesian:
Problem Statement:
A new credit scoring system has been developed to predict the
likelihood of loan defaults. The system has a 90% sensitivity, meaning
that it correctly identifies 90% of those who will default on their
loans. It also has a 95% specificity, meaning that it correctly
identifies 95% of those who will not default. The default rate among
borrowers is 2%. Given these prevalence, sensitivity, and specificity
estimates, what is the probability that a borrower flagged by the system
as likely to default will actually default? If the average loss per
defaulted loan is $200,000 and the cost to run the credit scoring test
on each borrower is $500, what is the total first-year cost for
evaluating 10,000 borrowers?
Solution:
A credit scoring system has been developed with the following
metrics:
(\(P(\text{Positive Test} \mid
\text{Default})\)) = 90%
(\(P(\text{Negative Test} \mid
\text{No Default})\)) = 95%
(\(P(\text{Default})\)) =
2%
We need to calculate:
The probability that a borrower flagged as likely to default will
actually default.
The total cost of evaluating 10,000 borrowers, where:
- The average loss per defaulted loan is $200,000.
- The cost to run the test for each borrower is $500.
Using Bayes’ Theorem, we have:
\[
P(\text{Default} \mid \text{Positive Test}) = \frac{P(\text{Positive
Test} \mid \text{Default}) \cdot P(\text{Default})}{P(\text{Positive
Test})}
\]
Where: \[
P(\text{Positive Test}) = P(\text{Positive Test} \mid \text{Default})
\cdot P(\text{Default}) + P(\text{Positive Test} \mid \text{No Default})
\cdot P(\text{No Default})
\]
Given: \[
P(\text{Positive Test} \mid \text{Default}) = 0.90, \quad
P(\text{Default}) = 0.02
\] \[
P(\text{Positive Test} \mid \text{No Default}) = 1 - \text{Specificity}
= 1 - 0.95 = 0.05
\] \[
P(\text{No Default}) = 1 - P(\text{Default}) = 1 - 0.02 = 0.98
\]
Substitute these values:
\[
P(\text{Positive Test}) = (0.90 \times 0.02) + (0.05 \times 0.98) =
0.018 + 0.049 = 0.067
\]
Thus:
\[
P(\text{Default} \mid \text{Positive Test}) = \frac{0.90 \times
0.02}{0.067} \approx 0.2687 \text{ or } 26.87\%
\]
Given: - Average loss per defaulted loan = $200,000 - Cost of test
per borrower = $500 - Number of borrowers = 10,000
First, calculate the total test cost:
\[
\text{Total Test Cost} = 10,000 \times 500 = 5,000,000
\]
Then, calculate the expected loss cost:
\[
\text{Expected Defaults in Flagged Positives} = P(\text{Default} \mid
\text{Positive Test}) \times 10,000 \times 0.02
\] \[
= 0.2687 \times 10,000 \times 0.02 = 53.74
\]
\[
\text{Expected Loss Cost} = 53.74 \times 200,000 \approx 10,747,462.69
\]
So, the total cost for the first year is:
\[
\text{Total First-Year Cost} = \text{Total Test Cost} + \text{Expected
Loss Cost} = 5,000,000 + 10,747,462.69 = 15,747,462.69
\]
Here is the same solution using R Language:
# Given values
sensitivity <- 0.90 # P(Test Positive | Default)
specificity <- 0.95 # P(Test Negative | No Default)
prevalence <- 0.02 # P(Default)
non_default_prevalence <- 1 - prevalence # P(No Default)
# Calculate P(Test Positive)
p_test_positive <- (sensitivity * prevalence) + ((1 - specificity) * non_default_prevalence)
# Calculate P(Default | Test Positive) using Bayes' Theorem
ppv <- (sensitivity * prevalence) / p_test_positive
ppv
## [1] 0.2686567
# Cost values
loss_per_default <- 200000 # Average loss per defaulted loan
test_cost_per_borrower <- 500 # Cost to run the credit scoring test on each borrower
num_borrowers <- 10000 # Number of borrowers
# Expected defaults in flagged positives (PPV * flagged positives)
expected_defaults_in_flagged <- ppv * num_borrowers * prevalence
# Total cost: Test cost for all borrowers + Expected losses from defaults
total_test_cost <- num_borrowers * test_cost_per_borrower
expected_loss_cost <- expected_defaults_in_flagged * loss_per_default
total_first_year_cost <- total_test_cost + expected_loss_cost
total_first_year_cost
## [1] 15746269
Binomial:
Problem Statement:
The probability that a stock will pay a dividend in any given quarter
is 0.7. What is the probability that the stock pays dividends exactly 6
times in 8 quarters? What is the probability that it pays dividends 6 or
more times? What is the probability that it pays dividends fewer than 6
times? What is the expected number of dividend payments over 8 quarters?
What is the standard deviation?
Solution:
The probability that a stock will pay a dividend in any given quarter
is 0.7. Over 8 quarters, we need to calculate:
The probability that the stock pays dividends exactly 6
times.
The probability that it pays dividends 6 or more times.
The probability that it pays dividends fewer than 6
times.
The expected number of dividend payments over 8
quarters.
The standard deviation of dividend payments.
The dividend payments follow a Binomial distribution
where: \[
n = 8 \quad \text{and} \quad p = 0.7
\]
Using the Binomial probability formula: \[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
Given \(n = 8\), \(k = 6\), and \(p
= 0.7\), we find \(P(X =
6)\):
\[
P(X = 6) = \binom{8}{6} (0.7)^6 (0.3)^2
\] \[
= \frac{8!}{6!(8-6)!} \times (0.7)^6 \times (0.3)^2
\] \[
\approx 0.2965
\]
Thus, the probability that the stock pays dividends exactly 6 times
in 8 quarters is approximately 0.2965.
To find \(P(X \geq 6)\), we sum the
probabilities for 6, 7, and 8 dividends: \[
P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8)
\]
Using a binomial cumulative distribution: \[
P(X \geq 6) \approx 0.5518
\]
So, the probability that it pays dividends 6 or more times is
approximately 0.5518.
To find \(P(X < 6)\), we sum the
probabilities for 0 through 5 dividends: \[
P(X < 6) = P(X = 0) + P(X = 1) + \dots + P(X = 5)
\]
Using the cumulative distribution function: \[
P(X < 6) \approx 0.4482
\]
Thus, the probability that it pays dividends fewer than 6 times is
approximately 0.4482.
The expected number of successes \(E(X)\) in a Binomial distribution is given
by: \[
E(X) = n \cdot p
\]
Substitute \(n = 8\) and \(p = 0.7\): \[
E(X) = 8 \times 0.7 = 5.6
\]
So, the expected number of dividend payments over 8 quarters is
5.6.
The standard deviation \(\sigma\)
for a Binomial distribution is given by: \[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]
Substitute \(n = 8\) and \(p = 0.7\): \[
\sigma = \sqrt{8 \times 0.7 \times 0.3} \approx 1.2961
\]
Thus, the standard deviation of dividend payments over 8 quarters is
approximately 1.2961.
Here is the same solution using R language
# Given parameters
n <- 8 # Number of trials (quarters)
p <- 0.7 # Probability of a dividend payment per quarter
# 1. Probability of exactly 6 dividends in 8 quarters
k <- 6
prob_6_dividends <- dbinom(k, n, p)
cat("Probability of exactly 6 dividends in 8 quarters:", round(prob_6_dividends, 4), "\n")
## Probability of exactly 6 dividends in 8 quarters: 0.2965
# 2. Probability of 6 or more dividends in 8 quarters
# P(X >= 6) = P(X = 6) + P(X = 7) + P(X = 8)
prob_6_or_more_dividends <- pbinom(5, n, p, lower.tail = FALSE)
cat("Probability of 6 or more dividends in 8 quarters:", round(prob_6_or_more_dividends, 4), "\n")
## Probability of 6 or more dividends in 8 quarters: 0.5518
# 3. Probability of fewer than 6 dividends in 8 quarters
# P(X < 6) = P(X <= 5)
prob_fewer_than_6_dividends <- pbinom(5, n, p)
cat("Probability of fewer than 6 dividends in 8 quarters:", round(prob_fewer_than_6_dividends, 4), "\n")
## Probability of fewer than 6 dividends in 8 quarters: 0.4482
# 4. Expected number of dividend payments
expected_dividends <- n * p
cat("Expected number of dividend payments over 8 quarters:", expected_dividends, "\n")
## Expected number of dividend payments over 8 quarters: 5.6
# 5. Standard deviation of dividend payments
std_deviation_dividends <- sqrt(n * p * (1 - p))
cat("Standard deviation of dividend payments:", round(std_deviation_dividends, 4), "\n")
## Standard deviation of dividend payments: 1.2961
Poisson:
Problem Statement:
A financial analyst notices that there are an average of 12 trading
days each month when a certain stock’s price increases by more than 2%.
What is the probability that exactly 4 such days occur in a given month?
What is the probability that more than 12 such days occur in a given
month? How many such days would you expect in a 6-month period? What is
the standard deviation of the number of such days? If an investment
strategy requires at least 70 days of such price increases in a year for
profitability, what is the percent utilization and what are your
recommendations?
Solution:
A financial analyst observes that there are, on average, 12 trading
days each month when a certain stock’s price increases by more than 2%.
We need to calculate:
- The probability that exactly 4 such days occur in a given
month.
- The probability that more than 12 such days occur in a given
month.
- The expected number of such days in a 6-month period.
- The standard deviation of the number of such days.
- If an investment strategy requires at least 70 days of such price
increases in a year for profitability, calculate the percent utilization
and provide recommendations.
We assume that the number of days follows a Poisson
distribution with a mean \(\lambda =
12\) (days per month).
Using the Poisson probability formula: \[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\]
Given \(\lambda = 12\) and \(k = 4\), we find \(P(X = 4)\): \[
P(X = 4) = \frac{12^4 e^{-12}}{4!}
\] \[
\approx 0.0027
\]
So, the probability of exactly 4 days in a month with a price
increase of over 2% is approximately 0.27%.
To find \(P(X > 12)\), we use:
\[
P(X > 12) = 1 - P(X \leq 12)
\]
Using cumulative probability for the Poisson distribution: \[
P(X > 12) \approx 0.384
\]
Therefore, the probability of more than 12 such days is approximately
38.4%.
The expected value for a Poisson distribution over 6 months is: \[
E(X) = 6 \cdot \lambda = 6 \cdot 12 = 72
\]
Thus, we expect approximately 72 days over a 6-month period.
The standard deviation \(\sigma\)
for a Poisson distribution is given by: \[
\sigma = \sqrt{\lambda}
\]
For a 6-month period: \[
\sigma = \sqrt{6 \cdot 12} = \sqrt{72} \approx 8.49
\]
So, the standard deviation over 6 months is approximately 8.49
days.
If an investment strategy requires at least 70 days of such price
increases in a year, the expected number of days is: \[
E(X) = 12 \cdot 12 = 144 \quad \text{(days per year)}
\]
The percent utilization is: \[
\text{Utilization} = \frac{70}{144} \times 100 \approx 48.6\%
\]
Recommendation: Given that the percent utilization
is approximately 48.6%, the strategy is likely feasible if the target is
70 days, as the expected value is well above this threshold. However,
due to variability, further analysis on monthly variances or a
contingency plan is advised.
Here is the same solution using R-Code:
# Given parameters
lambda <- 12 # Average number of days per month with price increase > 2%
# 1. Probability of exactly 4 days in a month
k <- 4
prob_4_days <- dpois(k, lambda)
cat("Probability of exactly 4 days in a month:", round(prob_4_days, 4), "\n")
## Probability of exactly 4 days in a month: 0.0053
# 2. Probability of more than 12 days in a month
prob_more_than_12 <- ppois(12, lambda, lower.tail = FALSE)
cat("Probability of more than 12 days in a month:", round(prob_more_than_12, 4), "\n")
## Probability of more than 12 days in a month: 0.424
# 3. Expected number of days in a 6-month period
expected_6_months <- 6 * lambda
cat("Expected number of days in a 6-month period:", expected_6_months, "\n")
## Expected number of days in a 6-month period: 72
# 4. Standard deviation of the number of days in a 6-month period
std_dev_6_months <- sqrt(expected_6_months)
cat("Standard deviation of days over 6 months:", round(std_dev_6_months, 4), "\n")
## Standard deviation of days over 6 months: 8.4853
# 5. Percent utilization and recommendation
# Expected number of days in a year
expected_year <- 12 * lambda
target_days <- 70
percent_utilization <- (target_days / expected_year) * 100
cat("Percent utilization for 70 target days:", round(percent_utilization, 2), "%\n")
## Percent utilization for 70 target days: 48.61 %
Hypergeometric:
Problem Statement:
A hedge fund has a portfolio of 25 stocks, with 15 categorized as
high-risk and 10 as low-risk. The fund manager randomly selects 7 stocks
to closely monitor. If the manager selected 5 high-risk stocks and 2
low-risk stocks, what is the probability of selecting exactly 5
high-risk stocks if the selection was random? How many high-risk and
low-risk stocks would you expect to be selected?
Solution:
A hedge fund has a portfolio of 25 stocks, with 15 categorized as
high-risk and 10 as low-risk. The fund manager randomly selects 7 stocks
to monitor. We are asked to determine:
The probability that exactly 5 of the selected stocks are
high-risk.
The expected number of high-risk and low-risk stocks in the
sample.
This scenario follows a Hypergeometric distribution,
where: - \(N = 25\) (total number of
stocks), - \(K = 15\) (number of
high-risk stocks), - \(n = 7\) (number
of stocks selected), - \(k = 5\)
(number of high-risk stocks chosen in the sample).
The probability of selecting exactly 5 high-risk stocks is given by
the Hypergeometric formula: \[
P(X = k) = \frac{\binom{K}{k} \cdot \binom{N - K}{n - k}}{\binom{N}{n}}
\]
Substitute \(N = 25\), \(K = 15\), \(n =
7\), and \(k = 5\): \[
P(X = 5) = \frac{\binom{15}{5} \cdot \binom{10}{2}}{\binom{25}{7}}
\]
Calculating each component: \[
\binom{15}{5} = \frac{15!}{5!(15 - 5)!} = 3003, \quad \binom{10}{2} =
45, \quad \binom{25}{7} = 480700
\]
Substitute these values: \[
P(X = 5) = \frac{3003 \times 45}{480700} \approx 0.2812
\]
Thus, the probability of selecting exactly 5 high-risk stocks is
approximately 0.2812.
For a Hypergeometric distribution, the expected number of high-risk
stocks in the sample is: \[
E(\text{high-risk}) = n \cdot \frac{K}{N} = 7 \cdot \frac{15}{25} = 4.2
\]
Similarly, the expected number of low-risk stocks is: \[
E(\text{low-risk}) = n \cdot \frac{N - K}{N} = 7 \cdot \frac{10}{25} =
2.8
\]
Therefore, we expect to select approximately 4.2 high-risk and 2.8
low-risk stocks.
Her is the same solution using R-Code:
# Given parameters
N <- 25 # Total number of stocks
K <- 15 # Number of high-risk stocks
n <- 7 # Number of stocks selected
k <- 5 # Number of high-risk stocks chosen
# 1. Probability of selecting exactly 5 high-risk stocks
prob_5_high_risk <- dhyper(k, K, N - K, n)
cat("Probability of exactly 5 high-risk stocks:", round(prob_5_high_risk, 4), "\n")
## Probability of exactly 5 high-risk stocks: 0.2811
# 2. Expected number of high-risk and low-risk stocks
expected_high_risk <- n * (K / N)
expected_low_risk <- n * ((N - K) / N)
cat("Expected number of high-risk stocks:", round(expected_high_risk, 2), "\n")
## Expected number of high-risk stocks: 4.2
cat("Expected number of low-risk stocks:", round(expected_low_risk, 2), "\n")
## Expected number of low-risk stocks: 2.8
Geometric:
Problem Statement:
The probability that a bond defaults in any given year is 0.5%. A
portfolio manager holds this bond for 10 years. What is the probability
that the bond will default during this period? What is the probability
that it will default in the next 15 years? What is the expected number
of years before the bond defaults? If the bond has already survived 10
years, what is the probability that it will default in the next 2
years?
Solution:
The probability that a bond defaults in any given year is 0.5%. A
portfolio manager holds this bond for a specified period. We are asked
to determine:
The probability that the bond will default within 10
years.
The probability that the bond will default within 15
years.
The expected number of years before the bond defaults.
The probability that the bond defaults within 2 years, given that
it has already survived for 10 years.
This scenario follows a Geometric distribution,
where: - \(p = 0.005\) (probability of
default each year).
For a Geometric distribution, the probability that the bond defaults
within \(n\) years is given by the
cumulative distribution function: \[
P(X \leq n) = 1 - (1 - p)^n
\]
For \(p = 0.005\) and \(n = 10\): \[
P(X \leq 10) = 1 - (1 - 0.005)^{10}
\] \[
= 1 - (0.995)^{10} \approx 0.0488
\]
Thus, the probability that the bond will default within 10 years is
approximately 4.88%.
Similarly, for \(n = 15\): \[
P(X \leq 15) = 1 - (1 - 0.005)^{15}
\] \[
= 1 - (0.995)^{15} \approx 0.0729
\]
The probability of default within 15 years is approximately
7.29%.
The expected number of years before default for a Geometric
distribution is given by: \[
E(X) = \frac{1}{p}
\]
Substituting \(p = 0.005\): \[
E(X) = \frac{1}{0.005} = 200
\]
Thus, the expected number of years before default is 200 years.
Given that the bond has already survived for 10 years, the
probability that it defaults within the next 2 years is: \[
P(X \leq 2) = 1 - (1 - p)^2
\]
For \(p = 0.005\): \[
P(X \leq 2) = 1 - (0.995)^2 \approx 0.009975
\]
Therefore, the probability that the bond defaults in the next 2
years, given it has survived 10 years, is approximately 0.9975%.
Here is the same solution using R-Code:
# Given parameters
p <- 0.005 # Probability of default in a given year
# 1. Probability of default within 10 years
n1 <- 10
prob_within_10_years <- 1 - (1 - p)^n1
cat("Probability of default within 10 years:", round(prob_within_10_years, 4), "\n")
## Probability of default within 10 years: 0.0489
# 2. Probability of default within 15 years
n2 <- 15
prob_within_15_years <- 1 - (1 - p)^n2
cat("Probability of default within 15 years:", round(prob_within_15_years, 4), "\n")
## Probability of default within 15 years: 0.0724
# 3. Expected number of years before default
expected_years_before_default <- 1 / p
cat("Expected number of years before default:", expected_years_before_default, "\n")
## Expected number of years before default: 200
# 4. Probability of default within 2 years given survival for 10 years
n3 <- 2
prob_within_2_years_given_10_years <- 1 - (1 - p)^n3
cat("Probability of default within 2 years after surviving 10 years:", round(prob_within_2_years_given_10_years, 4), "\n")
## Probability of default within 2 years after surviving 10 years: 0.01
Poisson:
Problem Statement:
A high-frequency trading algorithm experiences a system failure about
once every 1500 trading hours. What is the probability that the
algorithm will experience more than two failures in 1500 hours? What is
the expected number of failures?
Solution:
A high-frequency trading algorithm experiences a system failure once
every 1500 trading hours on average. We need to calculate:
The probability that the algorithm will experience more than two
failures in 1500 hours.
The expected number of failures over 1500 hours.
Since this is a rare event over a fixed interval, we can model the
number of failures with a Poisson distribution. The
average rate (\(\lambda\)) for 1500
hours is 1 failure.
The expected number of failures in 1500 hours, given that failures
occur on average once every 1500 hours, is: \[
E(X) = \lambda = 1
\]
So, we expect 1 failure over 1500 hours.
For a Poisson distribution, the probability of observing more than
\(k\) events is calculated as: \[
P(X > k) = 1 - P(X \leq k)
\]
In this case, we need the probability of more than 2 failures (\(k = 2\)).
First, we calculate \(P(X \leq 2)\),
the cumulative probability for 0, 1, and 2 failures: \[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]
Each term in the Poisson probability mass function is given by: \[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\]
For \(\lambda = 1\): \[
P(X = 0) = \frac{1^0 e^{-1}}{0!} = e^{-1} \approx 0.3679
\] \[
P(X = 1) = \frac{1^1 e^{-1}}{1!} = e^{-1} \approx 0.3679
\] \[
P(X = 2) = \frac{1^2 e^{-1}}{2!} = \frac{e^{-1}}{2} \approx 0.1839
\]
Now, summing these probabilities: \[
P(X \leq 2) = 0.3679 + 0.3679 + 0.1839 = 0.9197
\]
Then, \[
P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9197 = 0.0803
\]
Thus, the probability of experiencing more than two failures in 1500
hours is approximately 8.03%.
\end{document}
R-Code:
# Given parameter
lambda <- 1 # Mean number of failures in 1500 hours
# 1. Expected number of failures
expected_failures <- lambda
cat("Expected number of failures in 1500 hours:", expected_failures, "\n")
## Expected number of failures in 1500 hours: 1
# 2. Probability of more than 2 failures
prob_more_than_2 <- 1 - ppois(2, lambda)
cat("Probability of more than 2 failures in 1500 hours:", round(prob_more_than_2, 4), "\n")
## Probability of more than 2 failures in 1500 hours: 0.0803
Uniform Distribution:
Problem Statement:
An investor is trying to time the market and is monitoring a stock
that they believe has an equal chance of reaching a target price between
20 and 60 days. What is the probability that the stock will reach the
target price in more than 40 days? If it hasn’t reached the target price
by day 40, what is the probability that it will reach it in the next 10
days? What is the expected time for the stock to reach the target
price?
Solution:
An investor monitors a stock expected to reach a target price between
20 and 60 days with equal probability for any day in this range. We are
asked to calculate:
The probability that the stock will reach the target price in
more than 40 days.
Given it hasn’t reached the target price by day 40, the
probability it will reach it in the next 10 days.
The expected time for the stock to reach the target
price.
Since the stock reaching the target price is uniformly distributed,
we can use the properties of the Uniform distribution
on the interval \([a, b]\).
For a continuous uniform distribution over \([a, b]\), the probability that the time
\(X\) is greater than a certain day
\(c\) is: \[
P(X > c) = \frac{b - c}{b - a}
\]
Substituting \(c = 40\), \(a = 20\), and \(b
= 60\): \[
P(X > 40) = \frac{60 - 40}{60 - 20} = \frac{20}{40} = 0.5
\]
So, the probability that the stock will reach the target price in
more than 40 days is 0.5 (or 50%).
Given that the target hasn’t been reached by day 40, we are
interested in the conditional probability that it reaches the target in
the next 10 days (between days 40 and 50). This is calculated as: \[
P(40 < X \leq 50 \mid X > 40) = \frac{P(40 < X \leq 50)}{P(X
> 40)}
\]
We already know \(P(X > 40) =
0.5\). Now we calculate \(P(40 < X
\leq 50)\) as: \[
P(40 < X \leq 50) = \frac{50 - 40}{60 - 20} = \frac{10}{40} = 0.25
\]
Thus, \[
P(40 < X \leq 50 \mid X > 40) = \frac{0.25}{0.5} = 0.5
\]
So, given it hasn’t reached by day 40, the probability of reaching
the target in the next 10 days is also 0.5 (or 50%).
The expected value \(E(X)\) for a
uniform distribution on \([a, b]\) is
given by: \[
E(X) = \frac{a + b}{2}
\]
Substituting \(a = 20\) and \(b = 60\): \[
E(X) = \frac{20 + 60}{2} = \frac{80}{2} = 40
\]
So, the expected time for the stock to reach the target price is 40
days.
R-Code:
# Given parameters
a <- 20 # Minimum time in days
b <- 60 # Maximum time in days
c <- 40 # Given day for probability calculations
# 1. Probability that the stock reaches the target price in more than 40 days
prob_more_than_40 <- (b - c) / (b - a)
cat("Probability of reaching the target price in more than 40 days:", prob_more_than_40, "\n")
## Probability of reaching the target price in more than 40 days: 0.5
# 2. Probability that the stock reaches the target in the next 10 days given it hasn’t by day 40
prob_40_to_50 <- (50 - 40) / (b - a) # Probability of reaching between day 40 and 50
cond_prob_40_to_50 <- prob_40_to_50 / prob_more_than_40
cat("Conditional probability of reaching target between days 40 and 50:", cond_prob_40_to_50, "\n")
## Conditional probability of reaching target between days 40 and 50: 0.5
# 3. Expected time to reach the target price
expected_time <- (a + b) / 2
cat("Expected time to reach the target price:", expected_time, "\n")
## Expected time to reach the target price: 40
Exponential Distribution:
Problem Statement:
A financial model estimates that the lifetime of a successful
start-up before it either goes public or fails follows an exponential
distribution with an expected value of 8 years. What is the expected
time until the start-up either goes public or fails? What is the
standard deviation? What is the probability that the start-up will go
public or fail after 6 years? Given that the start-up has survived for 6
years, what is the probability that it will go public or fail in the
next 2 years?
Solution:
A financial model estimates that the lifetime of a successful
start-up before it either goes public or fails follows an Exponential
distribution with an expected lifetime of 8 years. We need to
calculate:
The expected time until the start-up either goes public or
fails.
The standard deviation of the lifetime.
The probability that the start-up will go public or fail after 6
years.
Given that the start-up has survived for 6 years, the probability
it will go public or fail in the next 2 years.
For an Exponential distribution, the expected value \(E(X)\) is given as: \[
E(X) = \frac{1}{\lambda}
\]
Since the expected lifetime is already provided as 8 years, we have:
\[
E(X) = 8
\]
The standard deviation of an Exponential distribution is also \(\frac{1}{\lambda}\), which is equal to the
expected value. Therefore: \[
\text{Standard Deviation} = 8
\]
For an Exponential distribution, the probability that the lifetime
\(X\) exceeds a specific time \(t\) is: \[
P(X > t) = e^{-\lambda t}
\]
Substituting \(\lambda = 0.125\) and
\(t = 6\): \[
P(X > 6) = e^{-0.125 \times 6} = e^{-0.75} \approx 0.4724
\]
So, the probability that the start-up will go public or fail after 6
years is approximately 0.4724.
This is a conditional probability calculation. For an Exponential
distribution, the memoryless property allows us to calculate it as:
\[
P(X > 6 \text{ and } X \leq 8 \mid X > 6) = P(X \leq 2) = 1 -
e^{-\lambda \times 2}
\]
Substituting \(\lambda = 0.125\):
\[
P(X \leq 2) = 1 - e^{-0.125 \times 2} = 1 - e^{-0.25} \approx 0.2212
\]
Thus, given that the start-up has already survived 6 years, there is
approximately a 22.12% probability that it will go public or fail in the
next 2 years.
R-Code:
# Given parameters
expected_lifetime <- 8 # Expected lifetime in years
lambda <- 1 / expected_lifetime # Rate parameter
# 1. Expected time until the start-up goes public or fails
expected_time <- expected_lifetime
cat("Expected time until going public or failure:", expected_time, "years\n")
## Expected time until going public or failure: 8 years
# 2. Standard deviation of the lifetime
std_dev <- 1 / lambda
cat("Standard deviation of the lifetime:", std_dev, "years\n")
## Standard deviation of the lifetime: 8 years
# 3. Probability that the start-up will go public or fail after 6 years
t1 <- 6
prob_after_6_years <- exp(-lambda * t1)
cat("Probability of going public or failing after 6 years:", round(prob_after_6_years, 4), "\n")
## Probability of going public or failing after 6 years: 0.4724
# 4. Probability of going public or failing in the next 2 years given survival for 6 years
t2 <- 2
prob_next_2_years_given_6 <- 1 - exp(-lambda * t2)
cat("Probability of going public or failing in the next 2 years given survival for 6 years:", round(prob_next_2_years_given_6, 4), "\n")
## Probability of going public or failing in the next 2 years given survival for 6 years: 0.2212
Problem 2:
Product Selection:
Problem Statement:
A company produces 5 different types of green pens and 7 different
types of red pens. The marketing team needs to create a new promotional
package that includes 5 pens. How many different ways can the package be
created if it contains fewer than 2 green pens?
Solution:
# Number of ways to select 5 red pens from 7
case1 <- choose(7, 5)
# Number of ways to select 1 green pen from 5 and 4 red pens from 7
case2 <- choose(5, 1) * choose(7, 4)
# Total number of ways
total_ways <- case1 + case2
total_ways
## [1] 196
Marketing Campaign Outcomes:
Problem Statement:
A marketing campaign involves three stages: first, a customer is sent
5 email offers; second, the customer is targeted with 2 different online
ads; and third, the customer is presented with 3 personalized product
recommendations. If the email offers, online ads, and product
recommendations are selected randomly, how many different possible
outcomes are there for the entire campaign?
Solution:
# Number of email offers
email_offers <- 5
# Number of online ads
online_ads <- 2
# Number of product recommendations
product_recommendations <- 3
# Total possible outcomes
total_outcomes <- email_offers * online_ads * product_recommendations
total_outcomes
## [1] 30
Product Defect Probability:
Problem Statement:
A quality control team draws 3 products from a batch without
replacement. What is the probability that at least one of the products
drawn is defective if the defect rate is known to be consistent? Express
your answer as a fraction or a decimal number rounded to four decimal
places.
Solution:
# Defect rate
p <- 0.1
# Probability that at least one product is defective
prob_at_least_one_defective <- 1 - (1 - p)^3
round(prob_at_least_one_defective, 4)
## [1] 0.271
Business Strategy Choices:
A business strategist is choosing potential projects to invest in,
focusing on 17 high-risk, high-reward projects and 14 low-risk,
steady-return projects.
Step 1: How many different combinations of 5 projects can the
strategist select?
Step 2: How many different combinations of 5 projects can the
strategist select if they want at least one low-risk project?
# Total projects
total_projects <- 31
# High-risk and low-risk projects
high_risk <- 17
low_risk <- 14
# Step 1: Total combinations of 5 projects from 31
total_combinations <- choose(total_projects, 5)
# Step 2: Combinations of 5 projects with only high-risk projects
high_risk_only <- choose(high_risk, 5)
# Combinations with at least one low-risk project
at_least_one_low_risk <- total_combinations - high_risk_only
# Output results
total_combinations
## [1] 169911
at_least_one_low_risk
## [1] 163723
Event Scheduling:
Problem Statement:
A business conference needs to schedule 9 different keynote sessions
from three different industries: technology, finance, and healthcare.
There are 4 potential technology sessions, 104 finance sessions, and 17
healthcare sessions to choose from. How many different schedules can be
made? Express your answer in scientific notation rounding to the
hundredths place.
Solution:
# Total options for technology, finance, and healthcare sessions
technology_options <- 4
finance_options <- 104
healthcare_options <- 17
# Total available options to select 9 sessions
total_options <- technology_options + finance_options + healthcare_options
# Calculate combinations of choosing 9 sessions from total_options and then arrange them (order matters)
number_of_schedules <- choose(total_options, 9) * factorial(9)
# Expressing the result in scientific notation rounded to the hundredths place
number_of_schedules_sci <- format(number_of_schedules, scientific = TRUE, digits = 2)
print(number_of_schedules_sci)
## [1] "5.5e+18"
Book Selection for Corporate Training
Problem Statement:
An HR manager needs to create a reading list for a corporate
leadership training program, which includes 13 books in total. The books
are categorized into 6 novels, 6 business case studies, 7 leadership
theory books, and 5 strategy books.
Solution:
Step 1: If the manager wants to include no more than 4 strategy
books, how many different reading schedules are possible? Express your
answer in scientific notation rounding to the hundredths place.
# Number of books in each category
novels <- 6
business_case_studies <- 6
leadership_theory <- 7
strategy <- 5
# Initialize total schedules
total_schedules <- 0
# Loop over each possible count of strategy books from 0 to 4
for (s in 0:4) {
# Calculate the number of ways to select `s` strategy books from 5
strategy_choices <- choose(strategy, s)
# Remaining books needed from other categories to reach 13 books
remaining_books_needed <- 13 - s
# Calculate the number of ways to select the remaining books from novels, business case studies, and leadership theory
remaining_choices <- choose(novels + business_case_studies + leadership_theory, remaining_books_needed)
# Multiply the choices for strategy books with the choices for remaining books
total_schedules <- total_schedules + (strategy_choices * remaining_choices)
}
# Display result in scientific notation, rounded to two decimal places
paste0("Answer: ", format(total_schedules, scientific = TRUE, digits = 3))
## [1] "Answer: 2.42e+06"
Step 2: If the manager wants to include all 6 business case studies,
how many different reading schedules are possible? Express your answer
in scientific notation rounding to the hundredths place.
# Fixed choice of all 6 business case studies
business_case_choices <- 1
# Calculate the number of ways to select 7 more books from novels, leadership theory, and strategy books
remaining_choices_step2 <- choose(novels + leadership_theory + strategy, 7)
# Total schedules when all 6 business case studies are included
total_schedules_step2 <- business_case_choices * remaining_choices_step2
# Display result in scientific notation, rounded to two decimal places
paste0("Answer: ", format(total_schedules_step2, scientific = TRUE, digits = 3))
## [1] "Answer: 3.18e+04"
Product Arrangement
Problem Statement:
A retailer is arranging 10 products on a display shelf. There are 5
different electronic gadgets and 5 different accessories. What is the
probability that all the gadgets are placed together and all the
accessories are placed together on the shelf? Express your answer as a
fraction or a decimal number rounded to four decimal places.
Solution:
# Factorials of the necessary numbers
total_arrangements <- factorial(10)
favorable_arrangements <- factorial(2) * factorial(5) * factorial(5)
# Probability calculation
probability <- favorable_arrangements / total_arrangements
round(probability, 4)
## [1] 0.0079
Expected Value of a Business Deal:
A company is evaluating a deal where they either gain $4 for every
successful contract or lose $16 for every unsuccessful contract. A
“successful” contract is defined as drawing a queen or lower from a
standard deck of cards. (Aces are considered the highest card in the
deck.)
Step 1: Find the expected value of the deal. Round your answer to two
decimal places. Losses must be expressed as negative values.
# Probabilities of success and failure
p_success <- 48 / 52
p_failure <- 4 / 52
# Gains and losses
gain_success <- 4
loss_failure <- -16
# Expected value calculation
expected_value <- (p_success * gain_success) + (p_failure * loss_failure)
round(expected_value, 2)
## [1] 2.46
Step 2: If the company enters into this deal 833 times, how much
would they expect to win or lose? Round your answer to two decimal
places. Losses must be expressed as negative values.
# Number of deals
num_deals <- 833
# Total expected outcome
total_expected_outcome <- expected_value * num_deals
round(total_expected_outcome, 2)
## [1] 2050.46
