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.
Formal Definition and Use of the Formulae
Introduction
Conditional probability describes the probability of an event occurring given that another event has already occurred. It is an essential concept in probability theory and real-world applications, including medical studies and risk assessments.
Key Concepts
1. Conditional Probability Formula
- The probability of event \( A \) occurring given that event \( B \) has already occurred is given by:
\( P(A|B) = \frac{P(A \cap B)}{P(B)} \)
- \( P(A|B) \) represents the conditional probability of \( A \) given \( B \).
- \( P(A \cap B) \) is the probability of both \( A \) and \( B \) occurring.
- \( P(B) \) is the probability of event \( B \) occurring.
2. Independent Events
- Two events \( A \) and \( B \) are independent if knowing that \( B \) occurred does not affect the probability of \( A \).
- Mathematically, independence is expressed as:
\( P(A|B) = P(A) \) and \( P(A|B’) = P(A) \)
- Alternatively, the formula can be written as:
\( P(A \cap B) = P(A) P(B) \)
3. Testing for Independence
- To check whether two events are independent, verify if:
\( P(A \cap B) = P(A) P(B) \)
- If this condition holds, \( A \) and \( B \) are independent; otherwise, they are dependent.
Guidance, Clarification, and Syllabus Links
- Students should understand both the formal definitions and practical applications of conditional probability.
- Be able to recognize and test for independence in probability problems.
Solved Example
Problem: In a school, 40% of students play football (\( P(F) = 0.4 \)), and 25% of students play both football and basketball (\( P(F \cap B) = 0.25 \)). Find the probability that a randomly selected student plays basketball given that they already play football.
Step 1: Identify Given Values
- \( P(F) = 0.4 \)
- \( P(F \cap B) = 0.25 \)
Step 2: Apply the Conditional Probability Formula
\( P(B|F) = \frac{P(F \cap B)}{P(F)} \)
\( = \frac{0.25}{0.4} \)
\( = 0.625 \)
Step 3: Interpretation
The probability that a student plays basketball given that they play football is 0.625 (or 62.5%).
IB Mathematics AA SL Formal definition and use of the formulae Exam Style Worked Out Questions
Question
Events A and B are such that \({\rm{P}}(A) = 0.3\) , \({\rm{P}}(B) = 0.6\) and \({\rm{P}}(A \cup B) = 0.7\) .
The values q , r , s and t represent probabilities.
Write down the value of t .
(i) Show that \(r = 0.2\) .
(ii) Write down the value of q and of s .
(i) Write down \({\rm{P}}(B’)\) .
(ii) Find \({\rm{P}}(A|B’)\) .
Answer/Explanation
Markscheme
\(t = 0.3\) A1 N1
[1 mark]
(i) correct values A1
e.g. \(0.3 + 0.6 – 0.7\) , \(0.9 – 0.7\)
\(r = 0.2\) AG N0
(ii) \(q = 0.1\) , \(s = 0.4\) A1A1 N2
[3 marks]
(i) \(0.4\) A1 N1
(ii) \({\rm{P}}(A|B’) = \frac{1}{4}\) A2 N2
[3 marks]
Question
In a group of 20 girls, 13 take history and 8 take economics. Three girls take both history and economics, as shown in the following Venn diagram. The values \(p\) and \(q\) represent numbers of girls.
Find the value of \(p\);
Find the value of \(q\).
A girl is selected at random. Find the probability that she takes economics but not history.
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)\(p + 3 = 13,{\text{ }}13 – 3\)
\(p = 10\) A1 N2
[2 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(p + 3 + 5 + q = 20,{\text{ }}10 – 10 – 8\)
\(q = 2\) A1 N2
[2 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(20 – p – q – 3,{\text{ }}1 – \frac{{15}}{{20}},{\text{ }}n(E \cap H’) = 5\)
\(\frac{5}{{20}}\,\,\,\left( {\frac{1}{4}} \right)\) A1 N2
[2 marks]