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\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)
International Baccalaureate · DP 2: Analysis and Approaches HL

Paper 3 Workshop

⌚ ~27 min ▢ 3 blocks ✓ 10 exercises ➣ Prerequisites : Revision: Mixed Practice, Differential Equations
I Toolbox: The Law of Total Probability
Recappreviously studied — click to reread
Theorem — Law of Total Probability
For events \(E\) and \(F\) with \(0 < P(E) < 1\): $$P(F) = P(E) \PCond{F}{E} + P(E') \PCond{F}{E'}.$$ This applies when \(E\) and its complement \(E'\) form a partition of the sample space. More generally, if \((E_1,\dots,E_n)\) is a partition of the sample space into events of non-zero probability, then $$P(F) = \sum_{i=1}^n P(E_i)\PCond{F}{E_i}.$$
Method — Finding \(P(F)\) in a Tree
  1. Identify all paths to \(F\).
  2. Multiply probabilities along each path and sum them.
$$P(F) = \textcolor{colordef}{P(E) \PCond{F}{E}} + \textcolor{colorprop}{P(E') \PCond{F}{E'}}.$$
Figure
Example
For this probability tree,
Figure
find \(P(G)\).

  1. Paths to \(G\):
    Figure
  2. Calculate: $$\begin{aligned} P(G) &= \textcolor{colordef}{\frac{6}{10} \times \frac{1}{2}} + \textcolor{colorprop}{\frac{4}{10} \times \frac{2}{3}} \\ &= \textcolor{colordef}{\frac{3}{10}} + \textcolor{colorprop}{\frac{8}{30}} \\ &= \textcolor{colordef}{\frac{9}{30}} + \textcolor{colorprop}{\frac{8}{30}} \\ &= \frac{17}{30}. \end{aligned}$$

Skills to practice
  • Calculating Probabilities with Trees
II Toolbox: Solving by Separation of Variables
Recappreviously studied — click to reread
Method — Solving Separable Equations \(\frac{dy}{dx} \equal g(x)h(y)\)
This method applies when the equation can be rearranged so that all terms involving \(y\) are on one side and all terms involving \(x\) are on the other.
  1. Separate: Rearrange the equation into the form \(\dfrac{1}{h(y)}\, \mathrm{d}y = g(x)\, \mathrm{d}x\).
  2. Integrate: Integrate both sides of the equation with respect to their respective variables: $$ \int \dfrac{1}{h(y)} \, \mathrm{d}y = \int g(x) \, \mathrm{d}x $$
  3. Solve for \(y\): If possible, solve the resulting equation for \(y\) to get an explicit solution.
Before dividing, check whether the constant solutions given by \(h(y) = 0\) satisfy the equation: dividing by \(h(y)\) assumes \(h(y) \neq 0\) and can lose them.
Example
Solve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -2xy^2\) with the initial condition \(y(1)=1\).

We rearrange and integrate both sides: $$ \begin{aligned} \dfrac{\mathrm{d}y}{\mathrm{d}x} &= -2xy^2 \\ \frac{1}{y^2} \, \mathrm{d}y &= -2x \, \mathrm{d}x \\ \int \frac{1}{y^2} \, \mathrm{d}y &= \int -2x \, \mathrm{d}x \\ \int y^{-2} \, \mathrm{d}y &= \int -2x \, \mathrm{d}x \\ -y^{-1} &= -x^2 + C \\ -\frac{1}{y} &= -x^2 + C \end{aligned} $$ This is the implicit form of the solutions other than \(y = 0\). Using the initial condition \(y(1)=1\): $$ -\dfrac{1}{1} = -(1)^2 + C \implies -1 = -1 + C \implies C = 0 $$ Substituting \(C=0\) gives the particular solution: $$ -\dfrac{1}{y} = -x^2 \implies y = \dfrac{1}{x^2} $$ This solution is defined on \((0, +\infty)\), the interval containing the initial point \(x = 1\). Note that the constant function \(y = 0\) (lost when dividing by \(y^2\)) is also a solution of the equation, but it does not satisfy the initial condition.

Skills to practice
  • Finding Particular Solutions by Separation
III Exam-Style Questions
Skills to practice
  • Investigating with Technology: Paper 3