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At the Bank of California, past data show that \(8 \%\) of all credit card holders default at some time in their lives. On one recent day, this bank issued 12 credit cards to new customers. Find the probability that of these 12 customers, eventually a. exactly 3 will default b. exactly 1 will default \(\mathbf{c}\). none will default

Short Answer

Expert verified
The probabilities that out of the 12 new credit card holders, exactly 3, exactly 1, and none will default are calculated using the binomial distribution with parameters n = 12 and p = 0.08, using the values k = 3, k = 1, and k = 0 respectively.

Step by step solution

01

Understanding The Problem

Define the variables. Let X be the random variable representing the number of customers who will default. X follows a binomial distribution with parameters n and p. Here, n=12 (the number of trials, i.e., the number of credit cards issued) and p=0.08 (the probability of success, i.e., the probability that a customer will default).
02

Formulating The Probability Function

The probability mass function of a binomial distribution is: \( P(X=k) = C(n, k) * (p^k) * ((1-p)^{n-k}) \), where \( C(n, k) \) is the combination function defined as \( C(n,k)= n! / [k!(n-k)!] \).
03

Applying The Formula - Case A

For \( k=3 \), compute the value of the mass function: \( P(X=3) = C(12, 3) * (0.08^3) * ((1-0.08)^{12-3}) \). To evaluate the probability, use the combination function \( C(12, 3) \) (which is the number ways to pick 3 out of 12), and evaluate the rest of the equation using the given values of p and n.
04

Applying The Formula - Case B

Similarly, for \( k=1 \), compute the value of the mass function: \( P(X=1) = C(12, 1) * (0.08^1) * ((1-0.08)^{12-1}) \), using the same method as in Step 3.
05

Applying The Formula - Case C

Finally, for \( k=0 \) (no one defaults, compute the value of the binomial distribution: \( P(X=0) = C(12, 0) * (0.08^0) * ((1-0.08)^{12-0}) \), again using the method from Step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Mass Function
The Probability Mass Function (PMF) is a key concept when dealing with discrete random variables like those in a binomial distribution. It describes the probability that a discrete random variable is exactly equal to some value. In our example, the random variable is the number of credit card holders that will default. Each outcome, like exactly 3 defaults, can be computed using the PMF.

In the context of a binomial distribution, the PMF is given by the formula: \[ P(X = k) = C(n, k) \cdot p^k \cdot (1-p)^{n-k} \]Here,
  • \( P(X = k) \) is the probability of exactly \( k \) successes (e.g., defaults),
  • \( C(n, k) \) is the number of combinations of \( n \) items taken \( k \) at a time,
  • \( p^k \) is the probability of \( k \) successes,
  • \( (1-p)^{n-k} \) is the probability of all other \( n-k \) outcomes being failures.
This function helps determine the likelihood of various outcomes in the exercise.
Combination Function
The Combination Function, often represented as \( C(n, k) \), is crucial in calculating probabilities in binomial distributions. It determines the number of ways to choose \( k \) successes (defaults, in our case) out of \( n \) trials (the credit cards issued).

Mathematically, it is represented as:\[C(n, k) = \frac{n!}{k! (n-k)!}\]where \(!\) denotes a factorial, a product of all positive integers up to that number.

For our example:
  • To find \( C(12, 3) \), we calculate \( \frac{12!}{3! \cdot 9!} \), determining the ways to choose 3 defaults out of 12 issued cards.
  • Similarly, \( C(12, 1) \) is calculated as \( \frac{12!}{1! \cdot 11!} \) for exactly 1 default.
  • \( C(12, 0) \) is \( \frac{12!}{0! \cdot 12!} \), representing the scenario where none default.
This function is a building block in evaluating probabilities for each scenario outlined in probability mass function calculations.
Random Variable
A Random Variable is a fundamental concept in probability and statistics. It is a variable whose possible values are outcomes of a random phenomenon. In this specific exercise, the random variable \( X \) represents the number of customers who default on their credit card debt.

Two main characteristics of a random variable:
  • Discrete: Values are countable as they come from a finite set. For example, the count of defaulting customers can be 0, 1, 2, up to 12 if all defaulted.
  • Continuous: Cannot be applied here but relevant when dealing with ranges and non-countable outcomes.
Binomial random variables like \( X \) have two possible outcomes for each trial – success (default) or failure (no default). Thus, the binomial distribution aptly models the situation, providing probabilities for each possible value of \( X \). This helps evaluate real-world scenarios, just as with the credit card defaults.
Probability of Default
The Probability of Default is a specific probability that informs us about the chances of a credit card holder failing to fulfill their debt obligations. In the exercise, this probability is given as \( 0.08 \) or 8%.

This particular probability is used as a 'success' probability in binomial distribution because we are counting how many defaults (successes) occur among the issued credit cards.
  • The probability mass function uses this as part of its calculation for each of the cases – 3 defaults, 1 default, and no defaults.
  • This probability must be constant for the binomial distribution to be applicable.
To calculate, this potential outcome probability is raised to the power corresponding to the number of defaults and then multiplied by the inverse probability for non-default cases. Understanding this concept allows us to anticipate the financial impacts related to customer defaults effectively.

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Most popular questions from this chapter

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