Conditional Probabilities and Weighted Trees

Pharmaceutical Study: Introductory Example

• A pharmaceutical laboratory tested two treatments, Medication A ($A$) and Medication B ($B$), on a study sample of $800$ patients suffering from a specific disease.

• Results of the study are summarized in the following contingency table:

• Probabilities calculated across the entire patient sample ($800$ patients):

  • Event $A$: "The patient received Medication A."
  • Event $G$: "The patient is cured."
  • Probability that a patient received Medication A:     $P(A) = \frac{455}{800} \approx 0,57 = 57\%$
  • Probability that a patient is cured:     $P(G) = \frac{674}{800} \approx 0,84 = 84\%$
  • Probability that a patient is cured AND was treated with Medication A:     $P(G \cap A) = \frac{383}{800} \approx 0,48 = 48\%$
  • Probability that a patient is NOT cured AND was treated with Medication A:     $P(\overline{G} \cap A) = \frac{72}{800} = 0,09 = 9\%$

• Conditional probabilities calculated within specific sub-groups:

  • Choosing a cured patient at random ($674$ patients total):
  • Probability that the patient took Medication A given that they are cured:       $P_G(A) = \frac{383}{674} \approx 0,57 = 57\%$
  • Choosing a patient treated with Medication B at random ($345$ patients total):
  • Probability that the patient is cured given that they took Medication B:       $P_B(G) = \frac{291}{345} \approx 0,84 = 84\%$

Definition of Conditional Probability

• Verbatim Definition: The conditional probability of $B$ given $A$ is the probability that event $B$ occurs given that event $A$ has occurred.

• Notation: $P_A(B)$

• General Formula:   $P_A(B) = \frac{P(A \cap B)}{P(A)}$   where $P(A) \neq 0$.

Weighted Trees and Rules of Calculation

• Experimental Setup (Ball Extraction Model):
  • A bag contains $50$ total balls:
  • $20$ red balls ($R$)
  • $30$ black balls ($\overline{R}$)
  • Each ball is inscribed with either "Gagné" (Won, $G$) or "Perdu" (Lost, $\overline{G}$).
  • Distribution of inscriptions:
  • Among the $20$ red balls, $15$ are marked "Gagné".
  • Among the $30$ black balls, $9$ are marked "Gagné".
  • Events:
  • $R$: "Draw a red ball"
  • $\overline{R}$: "Draw a black ball"
  • $G$: "Draw a ball marked Gagné"
  • $R \cap G$: "Draw a red ball marked Gagné"

• Rule 1 (Sum of Probabilities at a Node):

  • Statement: From a single node, the sum of the probabilities of all outgoing branches is equal to $1$.
  • Application to initial ball selection node:     $P(R) = \frac{20}{50} = \frac{2}{5} = 0,4$$0,4 + P(\overline{R}) = 1$$P(\overline{R}) = 1 - 0,4 = 0,6$

• Rule 2 (Probability of a Path):

  • Statement: To calculate the probability of a path, multiply the probabilities of the individual branches along that path.
  • Application for path leading to $R \cap G$:
  • Conditional probability of winning given a red ball:       $P_R(G) = \frac{15}{20} = 0,75$
  • Path probability calculation:       $P(R \cap G) = P(R) \times P_R(G) = 0,4 \times 0,75 = 0,3$
  • Application for path leading to $\overline{R} \cap G$:
  • Conditional probability of winning given a black ball:       $P_{\overline{R}}(G) = \frac{9}{30} = 0,3$
  • Path probability calculation:       $P(\overline{R} \cap G) = P(\overline{R}) \times P_{\overline{R}}(G) = 0,6 \times 0,3 = 0,18$

• Rule 3 (Law of Total Probability / Formule des probabilités totales):

  • Statement: The probability of an event associated with multiple paths in a probability tree is equal to the sum of the probabilities of each of those individual paths.
  • Calculation for the event "Drawing a winning ball" ($G$):
  • Event $G$ is composed of the disjoint paths $R \cap G$ and $\overline{R} \cap G$$P(G) = P(R \cap G) + P(\overline{R} \cap G)$$P(G) = 0,3 + 0,18 = 0,48$

Practical Application: Medical Testing in Cattle

• Problem Context (BAC S, Antilles-Guyane 2010):

  • An epidemic affects cattle. If diagnosed early, the animal can be cured; otherwise, the disease is fatal.
  • A diagnostic test is evaluated on a sample population where $2\%$ of animals carry the disease.
  • Test diagnostic performance metrics:
  • Given an animal carries the disease, the test is positive in $85\%$ of cases ($P_M(T) = 0,85$).
  • Given an animal is healthy (non-carrier), the test is negative in $95\%$ of cases ($P_{\overline{M}}(\overline{T}) = 0,95$).
  • Event definitions:
  • $M$: "Be a carrier of the disease"
  • $T$: "Have a positive test result"

• Question 1: Constructing the Weighted Probability Tree

  • Initial event probabilities:     $P(M) = 0,02$$P(\overline{M}) = 1 - 0,02 = 0,98$
  • Conditional probabilities given disease carrier status $M$:     $P_M(T) = 0,85$$P_M(\overline{T}) = 1 - 0,85 = 0,15$
  • Conditional probabilities given healthy status $\overline{M}$:     $P_{\overline{M}}(\overline{T}) = 0,95$$P_{\overline{M}}(T) = 1 - 0,95 = 0,05$

• Question 2: Overall Probability of a Positive Test Result

  • An animal chosen at random tests positive via two distinct path intersections: $M \cap T$ and $\overline{M} \cap T$
  • Calculating branch path probabilities using Rule 2:     $P(M \cap T) = P(M) \times P_M(T) = 0,02 \times 0,85 = 0,017$$P(\overline{M} \cap T) = P(\overline{M}) \times P_{\overline{M}}(T) = 0,98 \times 0,05 = 0,049$
  • Applying Rule 3 (Law of Total Probability):     $P(T) = P(M \cap T) + P(\overline{M} \cap T)$$P(T) = 0,017 + 0,049 = 0,066$
  • Conclusion: The probability that a randomly selected animal tests positive is $6,6\%$

• Question 3: Probability of Being Diseased Given a Positive Test

  • Formula for conditional probability:     $P_T(M) = \frac{P(T \cap M)}{P(T)}$
  • Substituting calculated numerical values:     $P_T(M) = \frac{0,02 \times 0,85}{0,066} = \frac{0,017}{0,066} \approx 0,26$
  • Diagnostic Evaluation:     The probability that a cattle beast is actually diseased given that its test returned positive is approximately $26\%$.     Despite relatively high test sensitivity ($85\%$) and high specificity ($95\%$), the low disease prevalence ($2\%$) leads to a high frequency of false positives relative to true positives. Consequently, a positive test result alone is not definitive proof of illness.