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The Binomial Probability Distribution
ADA University, School of Business
Information Communication Technologies Agency, Statistics Unit
2025-11-05
By the end of this lecture, you will be able to:
๐ Topics Covered Today
๐ Definition 1: Binomial Experiment
A binomial experiment possesses the following properties:
๐ง Think (1 minute): Consider the following financial scenario individually:
โA credit analyst reviews 10 loan applications from different borrowers. Each application is either approved or rejected based on credit scores. Historical data shows a 65% approval rate.โ
๐ฅ Pair (1 minute): Discuss with your neighbor - does this meet all five binomial experiment properties?
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๐ฃ๏ธ Share (1 minute): Groups share their conclusions with the class
๐ณ Example 1
A bankโs loan portfolio contains four independent corporate loans, each with a historical 5% probability of default over the next year. The risk manager is interested in \(Y\), the number of loans that will default. Is this a binomial experiment?
โ Solution
Letโs verify each property:
Conclusion: Yes, this is a binomial experiment! The bank can use binomial distribution to assess portfolio risk.
๐ Example 2
An economist surveys 10 randomly selected consumers from a large population where 40% prefer brand A over competitors. Let \(Y\) be the number who prefer brand A. Does this meet binomial requirements?
๐ก Important Remark
When sampling without replacement from a finite population, trials are technically dependent. However:
Rule of Thumb: If the sample size is less than 5% of the population (\(n < 0.05N\)), the dependency is negligible and we can approximate with the binomial distribution.
For large consumer markets, this condition is easily satisfied!
Which of the following is NOT a binomial experiment?
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๐ Definition 2: Binomial Distribution
A random variable \(Y\) has a binomial distribution based on \(n\) trials with success probability \(p\) if and only if:
\[ p(y) = \binom{n}{y}p^yq^{n-y}, \quad y=0,1,2,\dotsc, n \quad \text{and} \quad 0\le p \le 1 \]
where \(q = 1 - p\) and \(\binom{n}{y} = \frac{n!}{y!(n-y)!}\)
Notation: We write \(Y \sim \text{Binomial}(n, p)\) or \(Y \sim B(n, p)\)
๐ก Binomial Theorem
\[ (q+p)^n = \binom{n}{0}q^n+\binom{n}{1}pq^{n-1}+\binom{n}{2}p^2q^{n-2}+\cdots+\binom{n}{n}p^n \]
Since \(p + q = 1\), we have \((q+p)^n = 1\), confirming that \(\sum_{y=0}^{n}p(y) = 1\) โ
๐ Explore Binomial Distributions
Visit this interactive applet: Binomial Distribution Applet
Activity: Experiment with different parameter values:
Financial Interpretation: Observe how changing default probability \(p\) affects the distribution shape!
๐ง Think (1.5 min): Why do we multiply \(\binom{n}{y}\), \(p^y\), and \(q^{n-y}\) together? What does each component represent in a portfolio of risky assets?
๐ฅ Pair (1.5 min): Explain your reasoning to your partner. - What is \(\binom{n}{y}\)? (Number of ways to select which assets default) - What is \(p^y\)? (Probability those specific assets default) - What is \(q^{n-y}\)? (Probability remaining assets donโt default)
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๐ฃ๏ธ Share (1 min): Volunteer pairs share insights with the class
๐ผ Example 3
A venture capital fund invests in 5 startup companies. Historical data shows that 5% of startups fail within the first year. Find the probability that at least one startup fails.
๐ข Solution
Let \(Y\) = number of failed startups. Then \(Y \sim \text{Binomial}(n=5, p=0.05)\).
\[ P(\text{at least one fails}) = 1 - P(Y = 0) \]
\[ P(Y = 0) = \binom{5}{0}(0.05)^0(0.95)^5 = 1 \times 1 \times (0.95)^5 = 0.774 \]
Therefore: \[ P(Y \geq 1) = 1 - 0.774 = 0.226 \]
Interpretation: Thereโs a 22.6% chance that at least one startup will fail, despite the individual failure rate being only 5%. This demonstrates the importance of diversification!
๐ Example 4
A pharmaceutical company knows that 30% of experimental drugs pass clinical trials without their new technology. They claim a new screening method increases success. With the new method, 9 out of 10 drugs passed trials. If the method were worthless, whatโs \(P(Y \geq 9)\)?
๐ Solution Setup
Let \(Y\) = number of drugs passing trials. If method is worthless: \(p = 0.3\), \(n = 10\).
We need to find \(P(Y \geq 9) = P(Y = 9) + P(Y = 10)\)
Probability of exactly nine successes:
\[ P(Y = 9) = \binom{10}{9} (0.3)^9 (0.7)^1 = 10 \times (0.3)^9 \times 0.7 = 0.000138 \]
Probability of exactly ten successes:
\[ P(Y = 10) = \binom{10}{10} (0.3)^{10} (0.7)^0 = (0.3)^{10} = 0.000006 \]
Therefore: \[ P(Y \geq 9) = 0.000138 + 0.000006 = 0.000144 \]
๐ฏ Statistical Interpretation
If the screening method is ineffective, observing 9+ successes has probability 0.0144% (about 1 in 7,000!). This provides very strong evidence that the new screening method is effective. A pharmaceutical company would likely invest in this technology based on such results.
An investor makes 6 independent trades with 50% probability of profit on each. What is the probability of exactly 4 profitable trades?
Hint: Use \(n=6\), \(p=0.5\), \(y=4\) in the binomial formula.
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๐ Example 5
A bond portfolio contains bonds with 5% default probability. If the fund holds 20 bonds, find the probability that at least four bonds default.
๐ข Solution
With \(Y \sim \text{Binomial}(n=20, p=0.05)\):
\[ P(Y\geq 4) = 1 - P(Y\leq 3) = 1 - \sum_{y=0}^3p(y) \]
We can calculate each term or use cumulative binomial tables. Using statistical software or tables:
\(P(Y \leq 3) \approx 0.9841\)
Therefore: \[ P(Y \geq 4) = 1 - 0.9841 = 0.0159 \]
Risk Assessment: Only about 1.6% chance of having 4+ defaults, suggesting the portfolio is relatively safe under current assumptions.
๐ Cumulative Binomial Tables
Tabulated values represent cumulative probabilities: \[ P(Y \leq a) = \sum_{y=0}^{a}p(y) \]
Steps for \(n=20\), \(p=0.05\), finding \(P(Y \leq 3)\):
Modern Approach: Use statistical software for exact calculations:
R: pbinom(3, 20, 0.05) gives \(P(Y \leq 3)\)
Python: scipy.stats.binom.cdf(3, 20, 0.05)
Excel: =BINOM.DIST(3, 20, 0.05, TRUE)
๐ Theorem 1: Mean and Variance
Let \(Y\) be a binomial random variable based on \(n\) trials with success probability \(p\). Then:
\[ \mu = E(Y) = np \]
\[ \sigma^2 = V(Y) = npq = np(1-p) \]
\[ \sigma = SD(Y) = \sqrt{npq} \]
where \(q = 1 - p\).
๐ก Financial Interpretation
Scenario: A bankโs credit card portfolio has 1000 accounts, each with 2% monthly default probability.
Tasks (in groups of 3-4):
1. Calculate expected number of defaults per month (1 min)
2. Calculate the standard deviation (1 min)
3. If portfolio grows to 10,000 accounts, what happens to expected defaults and standard deviation? Does the default probability change? (2 min)
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4. Present your insights and discuss risk implications (1 min/group)
๐ Example 6
A market research firm surveys 20 randomly selected consumers about a new product. 6 respondents expressed interest in purchasing. Estimate \(p\), the true proportion of the population interested in the product.
๐ Solution: Maximum Likelihood Approach
\[ L(p) = P(Y = 6 | p) = \binom{20}{6} p^6 (1 - p)^{14} \]
Question: What value of \(p\) maximizes this likelihood?
To maximize \(L(p) = \binom{20}{6} p^6 (1 - p)^{14}\):
\[ \ln L(p) = \ln\binom{20}{6} + 6\ln(p) + 14\ln(1-p) \]
Differentiate with respect to \(p\):
\[ \frac{d\ln L}{dp} = \frac{6}{p} - \frac{14}{1-p} \]
Set the derivative equal to zero:
\[ \frac{6}{p} - \frac{14}{1-p} = 0 \]
Multiply both sides by \(p(1-p)\):
\[ 6(1-p) - 14p = 0 \]
\[ 6 - 6p - 14p = 0 \]
\[ 6 = 20p \]
\[ p = \frac{6}{20} = 0.3 \]
โ Final Solution
Check the second derivative: \[ \frac{d^2\ln L}{dp^2} = -\frac{6}{p^2} - \frac{14}{(1-p)^2} \]
This is always negative for \(0 < p < 1\), confirming a maximum at \(p = 6/20\).
Maximum Likelihood Estimate: \[ \boxed{\hat{p} = \frac{6}{20} = 0.30 = 30\%} \]
๐ฏ General Result
For binomial data, the Maximum Likelihood Estimator (MLE) is always: \[ \hat{p} = \frac{Y}{n} = \frac{\text{number of successes}}{\text{number of trials}} \]
This is simply the sample proportion - intuitive and optimal!
A loan portfolio has 25 loans, each with 40% default probability (independent). What are \(E(Y)\) and \(SD(Y)\)?
Hint: \(E(Y) = np\) and \(SD(Y) = \sqrt{npq}\)
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Scenario: A manufacturing companyโs historical defect rate is 2%. Todayโs inspection found 5 defects in a sample of 50 products.
๐ง Think (2 min): Is this evidence that the defect rate has increased? Under \(p = 0.02\), whatโs \(E(Y)\)?
๐ฅ Pair (2 min): Calculate (or estimate) \(P(Y \geq 5)\) when \(n=50\), \(p=0.02\). Use the fact that \(E(Y) = 1\). Is 5 unusually high?
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๐ฃ๏ธ Share (1 min): Present conclusions about whether quality has deteriorated
โ Key Takeaways
Five Properties of Binomial Experiments:
๐ข Fixed number of trials \(n\)
โ๏ธ Two outcomes: success or failure
๐ฏ Constant probability \(p\)
๐ Independent trials
๐ Count of successes \(Y\)
Essential Formulas: \[ p(y) = \binom{n}{y}p^yq^{n-y}, \quad E(Y) = np, \quad V(Y) = npq, \quad \hat{p}_{\text{MLE}} = \frac{Y}{n} \]
Applications: Credit risk, portfolio management, quality control, market research, insurance claims
๐ Homework Problems
Due: Next class meeting
๐ For Further Study
Interactive Tools:
๐ Khan Academy: Binomial Probability
๐บ StatQuest YouTube: Binomial Distribution
Statistical Software:
๐ R: dbinom(), pbinom(), qbinom(), rbinom()
๐ Python: scipy.stats.binom
๐ Excel: =BINOM.DIST()
Textbook Reading:
Chapter 3.4: The Binomial Distribution
Chapter 3.5: Applications and Examples
Review: Expected Value and Variance (Chapter 3.2)
๐ฌ Contact Information:
Samir Orujov
Assistant Professor
School of Business ADA University
๐ง Email: sorujov@ada.edu.az
๐ข Office: D312 โฐ Office Hours: By appointment
๐ Next Class:
Topic: Poisson Distribution
Reading: Chapter 3.6
Preparation: Review binomial concepts
โฐ Reminders:
โ Complete all homework problems
โ Review MLE concept thoroughly
โ Work hard
๐ฌ Open Discussion (5 minutes)
Feel free to ask questions about:
๐ Binomial distribution properties and assumptions
๐งฎ Calculating probabilities using formulas or tables
๐ Mean and variance interpretations
๐ฏ Maximum likelihood estimation methodology
๐ผ Applications to finance, economics, or your research
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Mathematical Statistics - The Binomial Probability Distribution