|
| 1 | +# /// script |
| 2 | +# requires-python = ">=3.10" |
| 3 | +# dependencies = [ |
| 4 | +# "marimo", |
| 5 | +# "matplotlib==3.10.0", |
| 6 | +# "matplotlib-venn==1.1.1", |
| 7 | +# "numpy==2.2.2", |
| 8 | +# ] |
| 9 | +# /// |
| 10 | + |
| 11 | +import marimo |
| 12 | + |
| 13 | +__generated_with = "0.11.4" |
| 14 | +app = marimo.App(width="medium", app_title="Conditional Probability") |
| 15 | + |
| 16 | + |
| 17 | +@app.cell |
| 18 | +def _(): |
| 19 | + import marimo as mo |
| 20 | + return (mo,) |
| 21 | + |
| 22 | + |
| 23 | +@app.cell(hide_code=True) |
| 24 | +def _(mo): |
| 25 | + mo.md( |
| 26 | + r""" |
| 27 | + # Conditional Probability |
| 28 | +
|
| 29 | + _This notebook is a computational companion to the book ["Probability for Computer Scientists"](https://chrispiech.github.io/probabilityForComputerScientists/en/part1/cond_prob/), by Stanford professor Chris Piech._ |
| 30 | +
|
| 31 | + In probability theory, we often want to update our beliefs when we receive new information. |
| 32 | + Conditional probability helps us formalize this process by calculating "_what is the chance of |
| 33 | + event $E$ happening given that we have already observed some other event $F$?_"[<sup>1</sup>](https://chrispiech.github.io/probabilityForComputerScientists/en/part1/cond_prob/) |
| 34 | +
|
| 35 | + When we condition on an event $F$: |
| 36 | +
|
| 37 | + - We enter the universe where $F$ has occurred |
| 38 | + - Only outcomes consistent with $F$ are possible |
| 39 | + - Our sample space reduces to $F$ |
| 40 | + """ |
| 41 | + ) |
| 42 | + return |
| 43 | + |
| 44 | + |
| 45 | +@app.cell(hide_code=True) |
| 46 | +def _(mo): |
| 47 | + mo.md( |
| 48 | + r""" |
| 49 | + ## Definition of Conditional Probability |
| 50 | +
|
| 51 | + The probability of event $E$ given that event $F$ has occurred is denoted as $P(E \mid F)$ and is defined as: |
| 52 | +
|
| 53 | + $$P(E \mid F) = \frac{P(E \cap F)}{P(F)}$$ |
| 54 | +
|
| 55 | + This formula tells us that the conditional probability is the probability of both events occurring |
| 56 | + divided by the probability of the conditioning event. |
| 57 | +
|
| 58 | + Let's start with a visual example. |
| 59 | + """ |
| 60 | + ) |
| 61 | + return |
| 62 | + |
| 63 | + |
| 64 | +@app.cell |
| 65 | +def _(): |
| 66 | + import matplotlib.pyplot as plt |
| 67 | + from matplotlib_venn import venn3 |
| 68 | + import numpy as np |
| 69 | + return np, plt, venn3 |
| 70 | + |
| 71 | + |
| 72 | +@app.cell(hide_code=True) |
| 73 | +def _(mo, plt, venn3): |
| 74 | + # Create figure with square boundaries |
| 75 | + plt.figure(figsize=(10, 3)) |
| 76 | + |
| 77 | + # Draw square sample space first |
| 78 | + rect = plt.Rectangle((-2, -2), 4, 4, fill=False, color="gray", linestyle="--") |
| 79 | + plt.gca().add_patch(rect) |
| 80 | + |
| 81 | + # Set the axis limits to show the full rectangle |
| 82 | + plt.xlim(-2.5, 2.5) |
| 83 | + plt.ylim(-2.5, 2.5) |
| 84 | + |
| 85 | + # Create Venn diagram showing E and F |
| 86 | + # For venn3, subsets order is: (100, 010, 110, 001, 101, 011, 111) |
| 87 | + # Representing: (A, B, AB, C, AC, BC, ABC) |
| 88 | + v = venn3(subsets=(30, 20, 10, 40, 0, 0, 0), set_labels=("E", "F", "Rest")) |
| 89 | + |
| 90 | + # Customize colors |
| 91 | + if v: |
| 92 | + for id in ["100", "010", "110", "001"]: |
| 93 | + if v.get_patch_by_id(id): |
| 94 | + if id == "100": |
| 95 | + v.get_patch_by_id(id).set_color("#ffcccc") # Light red for E |
| 96 | + elif id == "010": |
| 97 | + v.get_patch_by_id(id).set_color("#ccffcc") # Light green for F |
| 98 | + elif id == "110": |
| 99 | + v.get_patch_by_id(id).set_color( |
| 100 | + "#e6ffe6" |
| 101 | + ) # Lighter green for intersection |
| 102 | + elif id == "001": |
| 103 | + v.get_patch_by_id(id).set_color("white") # White for rest |
| 104 | + |
| 105 | + plt.title("Conditional Probability in Sample Space") |
| 106 | + |
| 107 | + # Remove ticks but keep the box visible |
| 108 | + plt.gca().set_yticks([]) |
| 109 | + plt.gca().set_xticks([]) |
| 110 | + plt.axis("on") |
| 111 | + |
| 112 | + # Add sample space annotation with arrow |
| 113 | + plt.annotate( |
| 114 | + "Sample Space (100)", |
| 115 | + xy=(-1.5, 1.5), |
| 116 | + xytext=(-2.2, 2), |
| 117 | + bbox=dict(boxstyle="round,pad=0.5", fc="white", ec="gray"), |
| 118 | + arrowprops=dict(arrowstyle="->"), |
| 119 | + ) |
| 120 | + |
| 121 | + # Add explanation |
| 122 | + explanation = mo.md(r""" |
| 123 | + ### Visual Intuition |
| 124 | +
|
| 125 | + In our sample space of 100 outcomes: |
| 126 | +
|
| 127 | + - Event $E$ occurs in 40 cases (red region: 30 + 10) |
| 128 | + - Event $F$ occurs in 30 cases (green region: 20 + 10) |
| 129 | + - Both events occur together in 10 cases (overlap) |
| 130 | + - Remaining cases: 40 (to complete sample space of 100) |
| 131 | +
|
| 132 | + When we condition on $F$: |
| 133 | + $$P(E \mid F) = \frac{P(E \cap F)}{P(F)} = \frac{10}{30} = \frac{1}{3} \approx 0.33$$ |
| 134 | +
|
| 135 | + This means: When we know $F$ has occurred (restricting ourselves to the green region), |
| 136 | + the probability of $E$ also occurring is $\frac{1}{3}$ - as 10 out of the 30 cases in the |
| 137 | + green region also belong to the red region. |
| 138 | + """) |
| 139 | + |
| 140 | + mo.vstack([mo.center(plt.gcf()), explanation]) |
| 141 | + return explanation, id, rect, v |
| 142 | + |
| 143 | + |
| 144 | +@app.cell(hide_code=True) |
| 145 | +def _(mo): |
| 146 | + mo.md( |
| 147 | + r"Next, here's a function that computes $P(E \mid F)$, given $P( E \cap F)$ and $P(F)$" |
| 148 | + ) |
| 149 | + return |
| 150 | + |
| 151 | + |
| 152 | +@app.cell |
| 153 | +def _(): |
| 154 | + def conditional_probability(p_intersection, p_condition): |
| 155 | + if p_condition == 0: |
| 156 | + raise ValueError("Cannot condition on an impossible event") |
| 157 | + if p_intersection > p_condition: |
| 158 | + raise ValueError("P(E∩F) cannot be greater than P(F)") |
| 159 | + |
| 160 | + return p_intersection / p_condition |
| 161 | + return (conditional_probability,) |
| 162 | + |
| 163 | + |
| 164 | +@app.cell |
| 165 | +def _(conditional_probability): |
| 166 | + # Example 1: Rolling a die |
| 167 | + # E: Rolling an even number (2,4,6) |
| 168 | + # F: Rolling a number greater than 3 (4,5,6) |
| 169 | + p_even_given_greater_than_3 = conditional_probability(2 / 6, 3 / 6) |
| 170 | + print("Example 1: Rolling a die") |
| 171 | + print(f"P(Even | >3) = {p_even_given_greater_than_3}") # Should be 2/3 |
| 172 | + return (p_even_given_greater_than_3,) |
| 173 | + |
| 174 | + |
| 175 | +@app.cell |
| 176 | +def _(conditional_probability): |
| 177 | + # Example 2: Cards |
| 178 | + # E: Drawing a Heart |
| 179 | + # F: Drawing a Face card (J,Q,K) |
| 180 | + p_heart_given_face = conditional_probability(3 / 52, 12 / 52) |
| 181 | + print("\nExample 2: Drawing cards") |
| 182 | + print(f"P(Heart | Face card) = {p_heart_given_face}") # Should be 1/4 |
| 183 | + return (p_heart_given_face,) |
| 184 | + |
| 185 | + |
| 186 | +@app.cell |
| 187 | +def _(conditional_probability): |
| 188 | + # Example 3: Student grades |
| 189 | + # E: Getting an A |
| 190 | + # F: Studying more than 3 hours |
| 191 | + p_a_given_study = conditional_probability(0.24, 0.40) |
| 192 | + print("\nExample 3: Student grades") |
| 193 | + print(f"P(A | Studied >3hrs) = {p_a_given_study}") # Should be 0.6 |
| 194 | + return (p_a_given_study,) |
| 195 | + |
| 196 | + |
| 197 | +@app.cell |
| 198 | +def _(conditional_probability): |
| 199 | + # Example 4: Weather |
| 200 | + # E: Raining |
| 201 | + # F: Cloudy |
| 202 | + p_rain_given_cloudy = conditional_probability(0.15, 0.30) |
| 203 | + print("\nExample 4: Weather") |
| 204 | + print(f"P(Rain | Cloudy) = {p_rain_given_cloudy}") # Should be 0.5 |
| 205 | + return (p_rain_given_cloudy,) |
| 206 | + |
| 207 | + |
| 208 | +@app.cell |
| 209 | +def _(conditional_probability): |
| 210 | + # Example 5: Error cases |
| 211 | + print("\nExample 5: Error cases") |
| 212 | + try: |
| 213 | + # Cannot condition on impossible event |
| 214 | + conditional_probability(0.5, 0) |
| 215 | + except ValueError as e: |
| 216 | + print(f"Error 1: {e}") |
| 217 | + |
| 218 | + try: |
| 219 | + # Intersection cannot be larger than condition |
| 220 | + conditional_probability(0.7, 0.5) |
| 221 | + except ValueError as e: |
| 222 | + print(f"Error 2: {e}") |
| 223 | + return |
| 224 | + |
| 225 | + |
| 226 | +@app.cell(hide_code=True) |
| 227 | +def _(mo): |
| 228 | + mo.md( |
| 229 | + r""" |
| 230 | + ## The Conditional Paradigm |
| 231 | +
|
| 232 | + When we condition on an event, we enter a new probability universe. In this universe: |
| 233 | +
|
| 234 | + 1. All probability axioms still hold |
| 235 | + 2. We must consistently condition on the same event |
| 236 | + 3. Our sample space becomes the conditioning event |
| 237 | +
|
| 238 | + Here's how our familiar probability rules look when conditioned on event $G$: |
| 239 | +
|
| 240 | + | Rule | Original | Conditioned on $G$ | |
| 241 | + |------|----------|-------------------| |
| 242 | + | Axiom 1 | $0 \leq P(E) \leq 1$ | $0 \leq P(E \mid G) \leq 1$ | |
| 243 | + | Axiom 2 | $P(S) = 1$ | $P(S \mid G) = 1$ | |
| 244 | + | Axiom 3* | $P(E \cup F) = P(E) + P(F)$ | $P(E \cup F \mid G) = P(E \mid G) + P(F \mid G)$ | |
| 245 | + | Complement | $P(E^C) = 1 - P(E)$ | $P(E^C \mid G) = 1 - P(E \mid G)$ | |
| 246 | +
|
| 247 | + *_For mutually exclusive events_ |
| 248 | + """ |
| 249 | + ) |
| 250 | + return |
| 251 | + |
| 252 | + |
| 253 | +@app.cell(hide_code=True) |
| 254 | +def _(mo): |
| 255 | + mo.md( |
| 256 | + r""" |
| 257 | + ## Multiple Conditions |
| 258 | +
|
| 259 | + We can condition on multiple events. The notation $P(E \mid F,G)$ means "_the probability of $E$ |
| 260 | + occurring, given that both $F$ and $G$ have occurred._" |
| 261 | +
|
| 262 | + The conditional probability formula still holds in the universe where $G$ has occurred: |
| 263 | +
|
| 264 | + $$P(E \mid F,G) = \frac{P(E \cap F \mid G)}{P(F \mid G)}$$ |
| 265 | +
|
| 266 | + This is a powerful extension that allows us to update our probabilities as we receive |
| 267 | + multiple pieces of information. |
| 268 | + """ |
| 269 | + ) |
| 270 | + return |
| 271 | + |
| 272 | + |
| 273 | +@app.cell |
| 274 | +def _(): |
| 275 | + def multiple_conditional_probability( |
| 276 | + p_intersection_all, p_intersection_conditions, p_condition |
| 277 | + ): |
| 278 | + """Calculate P(E|F,G) = P(E∩F|G)/P(F|G) = P(E∩F∩G)/P(F∩G)""" |
| 279 | + if p_condition == 0: |
| 280 | + raise ValueError("Cannot condition on an impossible event") |
| 281 | + if p_intersection_conditions == 0: |
| 282 | + raise ValueError( |
| 283 | + "Cannot condition on an impossible combination of events" |
| 284 | + ) |
| 285 | + if p_intersection_all > p_intersection_conditions: |
| 286 | + raise ValueError("P(E∩F∩G) cannot be greater than P(F∩G)") |
| 287 | + |
| 288 | + return p_intersection_all / p_intersection_conditions |
| 289 | + return (multiple_conditional_probability,) |
| 290 | + |
| 291 | + |
| 292 | +@app.cell |
| 293 | +def _(multiple_conditional_probability): |
| 294 | + # Example: College admissions |
| 295 | + # E: Getting admitted |
| 296 | + # F: High GPA |
| 297 | + # G: Good test scores |
| 298 | + |
| 299 | + # P(E∩F∩G) = P(Admitted ∩ HighGPA ∩ GoodScore) = 0.15 |
| 300 | + # P(F∩G) = P(HighGPA ∩ GoodScore) = 0.25 |
| 301 | + |
| 302 | + p_admit_given_both = multiple_conditional_probability(0.15, 0.25, 0.25) |
| 303 | + print("College Admissions Example:") |
| 304 | + print( |
| 305 | + f"P(Admitted | High GPA, Good Scores) = {p_admit_given_both}" |
| 306 | + ) # Should be 0.6 |
| 307 | + |
| 308 | + # Error case: impossible condition |
| 309 | + try: |
| 310 | + multiple_conditional_probability(0.3, 0.2, 0.2) |
| 311 | + except ValueError as e: |
| 312 | + print(f"\nError case: {e}") |
| 313 | + return (p_admit_given_both,) |
| 314 | + |
| 315 | + |
| 316 | +@app.cell(hide_code=True) |
| 317 | +def _(mo): |
| 318 | + mo.md( |
| 319 | + r""" |
| 320 | + ## 🤔 Test Your Understanding |
| 321 | +
|
| 322 | + Which of these statements about conditional probability are true? |
| 323 | +
|
| 324 | + <details> |
| 325 | + <summary>Knowing F occurred always decreases the probability of E</summary> |
| 326 | + ❌ False! Conditioning on F can either increase or decrease P(E), depending on how E and F are related. |
| 327 | + </details> |
| 328 | +
|
| 329 | + <details> |
| 330 | + <summary>P(E|F) represents entering a new probability universe where F has occurred</summary> |
| 331 | + ✅ True! We restrict ourselves to only the outcomes where F occurred, making F our new sample space. |
| 332 | + </details> |
| 333 | +
|
| 334 | + <details> |
| 335 | + <summary>If P(E|F) = P(E), then E and F must be the same event</summary> |
| 336 | + ❌ False! This actually means E and F are independent - knowing one doesn't affect the other. |
| 337 | + </details> |
| 338 | +
|
| 339 | + <details> |
| 340 | + <summary>P(E|F) can be calculated by dividing P(E∩F) by P(F)</summary> |
| 341 | + ✅ True! This is the fundamental definition of conditional probability. |
| 342 | + </details> |
| 343 | + """ |
| 344 | + ) |
| 345 | + return |
| 346 | + |
| 347 | + |
| 348 | +@app.cell(hide_code=True) |
| 349 | +def _(mo): |
| 350 | + mo.md( |
| 351 | + """ |
| 352 | + ## Summary |
| 353 | +
|
| 354 | + You've learned: |
| 355 | +
|
| 356 | + - How conditional probability updates our beliefs with new information |
| 357 | + - The formula $P(E \mid F) = P(E \cap F)/P(F)$ and its intuition |
| 358 | + - How probability rules work in conditional universes |
| 359 | + - How to handle multiple conditions |
| 360 | +
|
| 361 | + In the next lesson, we'll explore **independence** - when knowing about one event |
| 362 | + tells us nothing about another. |
| 363 | + """ |
| 364 | + ) |
| 365 | + return |
| 366 | + |
| 367 | + |
| 368 | +if __name__ == "__main__": |
| 369 | + app.run() |
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