91Ó°ÊÓ

Probability in court. In forensic science, DNA fragments found at the scene of a crime can be compared with DNA fragments from a suspected criminal to determine the probability that a match occurs by chance. Suppose that DNA fragment \(A\) is found in \(1 \%\) of the population, fragment \(B\) is found in \(4 \%\) of the population, and fragment \(C\) is found in \(2.5 \%\) of the population. (a) If the three fragments contain independent information, what is the probability that a suspect's DNA will match all three of these fragment characteristics by chance? (b) Some people believe such a fragment analysis is flawed because different DNA fragments do not represent independent properties. As before, suppose that fragment \(A\) occurs in \(1 \%\) of the population. But now suppose the conditional probability of \(B\), given \(A\), is \(p(B \mid A)=0.40\) rather than \(0.040\), and \(p(C \mid A)=0.25\) rather than \(0.025\). There is no additional information about any relationship between \(B\) and \(C\). What is the probability of a match now?

Step by step solution

01

Understand the problem statement

The problem involves determining the probability of a DNA match occurring by chance, given the frequencies of different DNA fragments in the population. There are two parts to the problem: (a) assuming all fragments are independent, and (b) assuming some dependencies between the fragments.

02

Calculating Independent Probability

(a) To find the probability that a suspect's DNA will match all three fragments independently, we multiply the probabilities of each fragment being found in the population. These probabilities are given as: Fragment A (P(A)) = 0.01 Fragment B (P(B)) = 0.04 Fragment C (P(C)) = 0.025

03

Multiplying the Independent Probabilities

The probability of all three fragments matching independently is: \[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \] Substituting the given values: \[ P(A \cap B \cap C) = 0.01 \times 0.04 \times 0.025 \] Calculate the result: \[ P(A \cap B \cap C) = 0.00001 \] Hence, the probability of all three fragments (A, B, and C) matching independently is \(0.00001\), or \(0.001\%\).

04

Understanding Conditional Probability

(b) In this part, we consider the probability of fragments B and C given that fragment A is already present. The probabilities are given as: \[ P(A) = 0.01 \] \[ P(B \mid A) = 0.40 \] \[ P(C \mid A) = 0.25 \] This means the probability of encountering fragment B given that A has already been encountered is 0.40, and similarly for C given A is 0.25.

05

Multiplying Conditional Probabilities

To find the probability of a match when dependencies are accounted for, we use: \[ P(A \cap B \cap C) = P(A) \times P(B \mid A) \times P(C \mid A) \] Substituting the given values: \[ P(A \cap B \cap C) = 0.01 \times 0.40 \times 0.25 \] Calculate the result: \[ P(A \cap B \cap C) = 0.01 \times 0.10 \] \[ P(A \cap B \cap C) = 0.001 \] Hence, the probability of a match now (considering dependencies) is \(0.001\), or \(0.1\%\).

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Key Concepts

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

Independent Probability

In forensic DNA analysis, we often assume that certain events or characteristics are independent, meaning the occurrence of one event does not affect the occurrence of another. This can simplify calculations and help us understand complex scenarios.

• Imagine DNA fragment A is found in 1% of the population (P(A) = 0.01).
• DNA fragment B is found in 4% of the population (P(B) = 0.04).
• DNA fragment C is found in 2.5% of the population (P(C) = 0.025).

If these fragments are truly independent, to find the probability that a suspect's DNA will match all three fragments by chance, we simply multiply these probabilities:

\[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) = 0.01 \times 0.04 \times 0.025 = 0.00001 \] So, the probability is 0.00001 or 0.001%. This demonstrates how unlikely it is for an individual's DNA to coincidentally match all three fragments if they are independent.

Conditional Probability

Conditional probability accounts for the fact that the occurrence of one event may affect the occurrence of another. This is crucial in realistic forensic DNA analyses where fragments may be related.

Suppose we know that fragment A occurs in 1% of the population (P(A) = 0.01). Now, instead of assuming independence:

• The probability of fragment B given that fragment A is present (P(B | A)) is 0.40.
• The probability of fragment C given that fragment A is present (P(C | A)) is 0.25.

To find the probability of a match when considering these dependencies, use:
\[ P(A \cap B \cap C) = P(A) \times P(B \mid A) \times P(C \mid A) = 0.01 \times 0.40 \times 0.25 = 0.001 \] Thus, the probability of a match considering dependencies is 0.001 or 0.1%, which shows how the dependencies change the result significantly.

DNA Fragment Analysis

DNA fragment analysis plays a crucial role in forensic science. It involves examining specific regions of DNA to identify unique characteristics that can help in criminal investigations.
Each DNA fragment corresponds to a specific allele and can occur with varying frequencies in different populations. This variation helps forensic scientists match DNA from crime scenes to potential suspects.

In our example:

• Fragment A occurs in 1% of the population.
• Fragment B occurs in 4% of the population.
• Fragment C occurs in 2.5% of the population.

By analyzing these fragments and understanding their probabilities (both independently and conditionally), forensic experts can accurately estimate the likelihood of a DNA match.
This process helps balance the weight of DNA evidence in court and aids in distinguishing true matches from coincidental ones.