import numpy as np, matplotlib.pyplot as plt
import matplotlib.patches as mpatches
rng = np.random.default_rng(31)
BLUE="#2563eb"; INDIGO="#4f46e5"; GREEN="#059669"; AMBER="#d97706"; PURPLE="#7c3aed"; PINK="#db2777"; INK="#1a2138"; GRID="#e6e9f2"
plt.rcParams.update({"figure.facecolor":"white","axes.facecolor":"white","figure.dpi":110,"font.size":11,
   "axes.edgecolor":GRID,"axes.grid":True,"grid.color":GRID,"axes.axisbelow":True,"axes.spines.top":False,
   "axes.spines.right":False,"axes.titlesize":12,"axes.titleweight":"bold","legend.frameon":False})
print("ready")

ready

hearts = 13; faces = 12; heart_faces = 3      # J,Q,K of hearts
p_h = hearts/52; p_f = faces/52; p_both = heart_faces/52
p_or = p_h + p_f - p_both
print(f"P(heart)            = 13/52 = {p_h:.3f}")
print(f"P(face)             = 12/52 = {p_f:.3f}")
print(f"P(heart AND face)   =  3/52 = {p_both:.3f}")
print(f"P(heart OR face)    = {p_h:.3f} + {p_f:.3f} - {p_both:.3f} = {p_or:.3f}")

# a Venn picture of the deck
fig, ax = plt.subplots(figsize=(6.4,3.8))
ax.add_patch(mpatches.Circle((0.40,0.5),0.30, color=BLUE, alpha=0.30))
ax.add_patch(mpatches.Circle((0.62,0.5),0.24, color=AMBER, alpha=0.30))
ax.text(0.28,0.5,"hearts\nonly\n10",ha="center",va="center",fontsize=10)
ax.text(0.51,0.5,"both\n3",ha="center",va="center",fontsize=10,fontweight="bold")
ax.text(0.72,0.5,"faces\nonly\n9",ha="center",va="center",fontsize=10)
ax.text(0.5,0.93,"neither: 30 cards",ha="center",fontsize=9,color="#8b94b3")
ax.set_xlim(0,1); ax.set_ylim(0.1,1); ax.axis("off"); ax.set_title("P(heart OR face) = 22/52: count each card once")
plt.tight_layout(); plt.show()

P(heart)            = 13/52 = 0.250
P(face)             = 12/52 = 0.231
P(heart AND face)   =  3/52 = 0.058
P(heart OR face)    = 0.250 + 0.231 - 0.058 = 0.423

# one die: events {roll a 1} and {roll a 6} cannot both occur
p1, p6 = 1/6, 1/6
print(f"P(1 or 6) = P(1) + P(6) = {p1:.3f} + {p6:.3f} = {p1+p6:.3f}   (no overlap to subtract)")
print(f"P(1 AND 6) = 0  -> the events are mutually exclusive")

rolls = rng.integers(1,7,200_000)
emp = np.mean((rolls==1) | (rolls==6))
print(f"empirical P(1 or 6) over 200k rolls = {emp:.3f}")

P(1 or 6) = P(1) + P(6) = 0.167 + 0.167 = 0.333   (no overlap to subtract)
P(1 AND 6) = 0  -> the events are mutually exclusive
empirical P(1 or 6) over 200k rolls = 0.331

# two dice: P(both even). Each die: P(even)=3/6
p_even = 3/6
p_both_even = p_even * p_even
print(f"P(both dice even) = P(even) x P(even) = {p_even} x {p_even} = {p_both_even:.3f}")

d = rng.integers(1,7,(300_000,2))
emp = np.mean((d[:,0]%2==0) & (d[:,1]%2==0))
print(f"empirical (300k rolls)              = {emp:.3f}")
print("\nGeneral form (any AND): P(A and B) = P(A) x P(B|A).")
print("When A and B are independent, P(B|A) = P(B), so it reduces to P(A) x P(B).")

P(both dice even) = P(even) x P(even) = 0.5 x 0.5 = 0.250
empirical (300k rolls)              = 0.251

General form (any AND): P(A and B) = P(A) x P(B|A).
When A and B are independent, P(B|A) = P(B), so it reduces to P(A) x P(B).

def is_independent(pA, pB, pAB): return np.isclose(pAB, pA*pB)

# Scenario 1 (one die): A={even}, B={odd}  -> mutually exclusive
pA, pB, pAB = 3/6, 3/6, 0.0
print("ONE DIE: A=even, B=odd")
print(f"  P(A and B) = {pAB}   P(A)xP(B) = {pA*pB:.3f}   independent? {is_independent(pA,pB,pAB)}")
print("  -> mutually exclusive, and therefore NOT independent (each makes the other impossible)\n")

# Scenario 2 (two dice): A={die1 even}, B={die2 is 6}  -> independent
pA, pB, pAB = 3/6, 1/6, (3/6)*(1/6)
print("TWO DICE: A=die1 even, B=die2 is 6")
print(f"  P(A and B) = {pAB:.3f}   P(A)xP(B) = {pA*pB:.3f}   independent? {is_independent(pA,pB,pAB)}")
print("  -> can both happen, and one does not affect the other: independent")

fig, ax = plt.subplots(figsize=(7,3.6))
labels=["A and B","P(A)xP(B)"]
ax.bar([0,1],[0.0,(3/6)*(3/6)],width=0.5,color=[PINK,GRID]); 
ax.bar([3,4],[(3/6)*(1/6),(3/6)*(1/6)],width=0.5,color=[GREEN,GRID])
ax.set_xticks([0.5,3.5]); ax.set_xticklabels(["mutually exclusive\n(even vs odd)","independent\n(die1 even, die2 six)"])
ax.set_ylabel("probability"); ax.set_title("independence test: does P(A and B) = P(A)xP(B)?")
ax.text(0.5,0.28,"0 ≠ 0.25\nNOT independent",ha="center",fontsize=8,color=PINK)
ax.text(3.5,0.12,"0.083 = 0.083\nindependent",ha="center",fontsize=8,color=GREEN)
plt.tight_layout(); plt.show()

ONE DIE: A=even, B=odd
  P(A and B) = 0.0   P(A)xP(B) = 0.250   independent? False
  -> mutually exclusive, and therefore NOT independent (each makes the other impossible)

TWO DICE: A=die1 even, B=die2 is 6
  P(A and B) = 0.083   P(A)xP(B) = 0.083   independent? True
  -> can both happen, and one does not affect the other: independent

ns = np.arange(1, 11)
p_at_least_one = 1 - (5/6)**ns          # at least one six in n rolls
for n in [1,2,4,6]:
    print(f"P(at least one six in {n} rolls) = 1 - (5/6)^{n} = {1-(5/6)**n:.3f}")

fig, ax = plt.subplots(figsize=(7,3.6))
ax.plot(ns, p_at_least_one, "o-", color=INDIGO)
ax.axhline(0.5, ls="--", color=PINK); ax.set_ylim(0,1)
ax.set_xlabel("number of rolls (n)"); ax.set_ylabel("P(at least one six)")
ax.set_title("P(at least one six) = 1 - (5/6)^n climbs fast"); plt.tight_layout(); plt.show()
print("\nIt passes 50% at n=4 -- a handy gambling fact: four rolls to favor at least one six.")

P(at least one six in 1 rolls) = 1 - (5/6)^1 = 0.167
P(at least one six in 2 rolls) = 1 - (5/6)^2 = 0.306
P(at least one six in 4 rolls) = 1 - (5/6)^4 = 0.518
P(at least one six in 6 rolls) = 1 - (5/6)^6 = 0.665

It passes 50% at n=4 -- a handy gambling fact: four rolls to favor at least one six.

⚙️ Setup¶