IB Mathematics AA Formal definition and use of the formulae Study Notes
IB Mathematics AA Formal definition and use of the formulae Study Notes
IB Mathematics AA Formal definition and use of the formulae Notes Offer a clear explanation of Formal definition and use of the formulae, its mean and variance, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Formal definition and use of the formulae, its mean and variance.
Conditional Probability and Testing for Independence
Conditional Probability and Testing for Independence
Conditional probability is the probability that event \( A \) occurs given that event \( B \) has occurred. It is denoted:
\( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad \text{(provided } P(B) > 0) \)
This tells us how the probability of \( A \) changes when we know that \( B \) has occurred.
An alternate form of this is:
\( P(A \cap B) = P(B) P(A \mid B) \)
Independence of Events
Events \( A \) and \( B \) are independent if the occurrence of one does not affect the probability of the other. Mathematically:
\( P(A \mid B) = P(A) \quad \text{or equivalently} \quad P(B \mid A) = P(B) \)
This also implies:
\( P(A \cap B) = P(A) P(B) \)
Testing for Independence
To test whether \( A \) and \( B \) are independent:
- Check if \( P(A \cap B) = P(A) P(B) \)
- Alternatively, check if \( P(A \mid B) = P(A) \)
If either condition holds, the events are independent.
Example:
In a group of students, 60% play football (F), 40% play basketball (B), and 25% play both sports. Are football and basketball playing independent?
▶️ Answer/Explanation
\( P(F) = 0.6 \), \( P(B) = 0.4 \), \( P(F \cap B) = 0.25 \)
Compute \( P(F) P(B) = 0.6 \times 0.4 = 0.24 \)
Since \( P(F \cap B) = 0.25 \neq 0.24 \), the events are not independent.
Conclusion: Football and basketball playing are not independent in this group.
Example:
A factory produces widgets from two machines: Machine A and Machine B. 60% of the widgets come from Machine A, 40% from Machine B.
5% of Machine A’s widgets are defective; 10% of Machine B’s widgets are defective.
(a) If a widget is chosen at random, what is the probability that it is defective?
(b) If a widget is found defective, what is the probability that it came from Machine B?
(c) Are the events “defective” and “from Machine B” independent?
▶️ Answer/Explanation
Let \( A \): from Machine A, \( B \): from Machine B, \( D \): defective
\( P(A) = 0.6 \), \( P(B) = 0.4 \)
\( P(D \mid A) = 0.05 \), \( P(D \mid B) = 0.10 \)
Find \( P(D) \)
\( P(D) = P(A)P(D \mid A) + P(B)P(D \mid B) \)
\( = 0.6 \times 0.05 + 0.4 \times 0.10 \)
\( = 0.03 + 0.04 = 0.07 \)
Find \( P(B \mid D) \)
\( P(B \mid D) = \frac{P(B \cap D)}{P(D)} \)
\( P(B \cap D) = P(B) P(D \mid B) = 0.4 \times 0.10 = 0.04 \)
\( P(B \mid D) = 0.04 / 0.07 \approx 0.571 \)
Test independence
If independent: \( P(B \cap D) = P(B) P(D) \)
\( P(B) P(D) = 0.4 \times 0.07 = 0.028 \)
\( P(B \cap D) = 0.04 \neq 0.028 \)
Conclusion: “Defective” and “from Machine B” are not independent.