Probability Problems for Interviews — Solved with Clear Logic

# === Basic Probability: Die Roll and Coin Toss Examples ===
# Run this to see the core formula in action with real numbers.

def probability(favourable_outcomes, total_outcomes):
    """Returns probability as a decimal and a readable fraction string."""
    if total_outcomes == 0:
        raise ValueError("Total outcomes cannot be zero — that's not a valid sample space.")
    prob = favourable_outcomes / total_outcomes
    return prob, f"{favourable_outcomes}/{total_outcomes}"

# --- Example 1: Rolling a fair six-sided die ---
# Event: getting an even number (2, 4, or 6)
total_die_faces = 6
even_faces = [2, 4, 6]          # these are our favourable outcomes
favourable_even = len(even_faces)

prob_decimal, prob_fraction = probability(favourable_even, total_die_faces)
print(f"P(rolling an even number) = {prob_fraction} = {prob_decimal:.4f}")

# --- Example 2: Drawing a card — Addition Rule (OR) ---
# Event: drawing a King OR a Heart from a standard 52-card deck
# Kings: 4 cards. Hearts: 13 cards. BUT King of Hearts is in both — avoid double-counting!
total_cards = 52
kings = 4
hearts = 13
king_of_hearts = 1              # the overlap between the two groups

# P(King OR Heart) = P(King) + P(Heart) - P(King AND Heart)
favourable_king_or_heart = kings + hearts - king_of_hearts  # = 16
prob_king_or_heart = favourable_king_or_heart / total_cards
print(f"\nP(King OR Heart) = {favourable_king_or_heart}/{total_cards} = {prob_king_or_heart:.4f}")

# --- Example 3: Tossing two fair coins — Multiplication Rule (AND) ---
# Event: getting Heads on both coins
# The coins are independent — first coin result does NOT affect the second
p_heads_coin1 = 1 / 2          # P(Heads) on a fair coin
p_heads_coin2 = 1 / 2          # same, independent event

p_both_heads = p_heads_coin1 * p_heads_coin2   # multiply because AND + independent
print(f"\nP(Heads AND Heads) = 1/2 × 1/2 = {p_both_heads:.4f}")

# Sanity check: list every possible outcome
all_outcomes = [(c1, c2) for c1 in ['H', 'T'] for c2 in ['H', 'T']]
favourable = [o for o in all_outcomes if o == ('H', 'H')]
print(f"All outcomes: {all_outcomes}")
print(f"Favourable outcomes: {favourable}")
print(f"P(HH) by counting = {len(favourable)}/{len(all_outcomes)} = {len(favourable)/len(all_outcomes):.4f}")

# === Replacement vs No-Replacement: Cards and Marbles ===

import math

# --- Scenario: Drawing 2 Aces from a 52-card deck ---

# WITH REPLACEMENT: deck is reset after each draw — draws are independent
p_first_ace_with = 4 / 52       # 4 aces in 52 cards
p_second_ace_with = 4 / 52     # deck is restored — still 4 aces in 52
p_two_aces_with_replacement = p_first_ace_with * p_second_ace_with
print("=== WITH REPLACEMENT ===")
print(f"P(1st Ace)  = 4/52 = {p_first_ace_with:.6f}")
print(f"P(2nd Ace)  = 4/52 = {p_second_ace_with:.6f}  (deck unchanged)")
print(f"P(Two Aces) = {p_two_aces_with_replacement:.6f}")

print()

# WITHOUT REPLACEMENT: first ace is removed — second draw sees 51 cards, 3 aces
p_first_ace_without = 4 / 52   # 4 aces in 52 cards
p_second_ace_without = 3 / 51  # 3 aces remain in 51 remaining cards — KEY CHANGE
p_two_aces_no_replacement = p_first_ace_without * p_second_ace_without
print("=== WITHOUT REPLACEMENT ===")
print(f"P(1st Ace)         = 4/52 = {p_first_ace_without:.6f}")
print(f"P(2nd Ace | 1st was Ace) = 3/51 = {p_second_ace_without:.6f}  (deck shrinks!)")
print(f"P(Two Aces)        = {p_two_aces_no_replacement:.6f}")

print()
print(f"Difference between methods: {abs(p_two_aces_with_replacement - p_two_aces_no_replacement):.6f}")
print("(Ignoring replacement makes you ~14% over-confident in this case)")

print()
# --- Marble Bag Problem (classic aptitude style) ---
# Bag contains: 5 red, 3 blue, 2 green marbles (10 total)
# Q: P(drawing red then blue) WITHOUT replacement?

total_marbles = 10
red_marbles = 5
blue_marbles = 3

p_red_first = red_marbles / total_marbles          # 5/10
p_blue_second_given_red = blue_marbles / (total_marbles - 1)  # 3/9 — one red is gone

p_red_then_blue = p_red_first * p_blue_second_given_red
print("=== MARBLE BAG (No Replacement) ===")
print(f"P(Red first)          = {red_marbles}/{total_marbles} = {p_red_first:.4f}")
print(f"P(Blue second | Red)  = {blue_marbles}/{total_marbles-1} = {p_blue_second_given_red:.4f}")
print(f"P(Red then Blue)      = {p_red_then_blue:.4f}")
print(f"As a fraction         = {red_marbles*blue_marbles}/{total_marbles*(total_marbles-1)} = {red_marbles*blue_marbles}/90")

# === Combinatorics in Probability: At-Least-One and Multi-Event Problems ===

import math

def combinations(n, r):
    """nCr — number of ways to choose r items from n (order doesn't matter)"""
    return math.comb(n, r)   # Python 3.8+ has this built-in

# ---------------------------------------------------------------
# PROBLEM 1: "At Least One" — Use the Complement Trick
# ---------------------------------------------------------------
# Q: A bag has 4 red and 6 blue marbles. Draw 3 without replacement.
#    What's P(at least one red marble)?

total_marbles = 10
red = 4
blue = 6
draw = 3

# Total ways to draw 3 from 10
total_ways = combinations(total_marbles, draw)   # 10C3 = 120

# P(at least one red) = 1 - P(NO red at all)
# 'No red' means all 3 are blue — choose 3 from 6 blue marbles
all_blue_ways = combinations(blue, draw)          # 6C3 = 20

p_no_red = all_blue_ways / total_ways
p_at_least_one_red = 1 - p_no_red               # the complement shortcut

print("=== PROBLEM 1: At Least One Red ===")
print(f"Total ways to draw 3 from 10 : 10C3 = {total_ways}")
print(f"Ways to draw 3 all-blue      : 6C3  = {all_blue_ways}")
print(f"P(no red)                    = {all_blue_ways}/{total_ways} = {p_no_red:.4f}")
print(f"P(at least one red)          = 1 - {p_no_red:.4f} = {p_at_least_one_red:.4f}")

print()

# ---------------------------------------------------------------
# PROBLEM 2: Exactly K Successes — Combinations in Numerator
# ---------------------------------------------------------------
# Q: From a group of 7 men and 5 women, a committee of 4 is formed randomly.
#    What's P(exactly 2 men and 2 women on the committee)?

total_people = 12   # 7 men + 5 women
men = 7
women = 5
committee_size = 4

# Total ways to form a committee of 4 from 12 people
total_committees = combinations(total_people, committee_size)   # 12C4 = 495

# Favourable: choose exactly 2 men from 7 AND exactly 2 women from 5
# These are independent choices for the two sub-groups
ways_to_pick_2_men   = combinations(men, 2)     # 7C2 = 21
ways_to_pick_2_women = combinations(women, 2)   # 5C2 = 10
favourable_committees = ways_to_pick_2_men * ways_to_pick_2_women  # 210

p_exactly_2_and_2 = favourable_committees / total_committees
print("=== PROBLEM 2: Exactly 2 Men and 2 Women ===")
print(f"Total committee arrangements : 12C4 = {total_committees}")
print(f"Ways to pick 2 men from 7    : 7C2  = {ways_to_pick_2_men}")
print(f"Ways to pick 2 women from 5  : 5C2  = {ways_to_pick_2_women}")
print(f"Favourable arrangements      : {ways_to_pick_2_men} × {ways_to_pick_2_women} = {favourable_committees}")
print(f"P(exactly 2 men, 2 women)    = {favourable_committees}/{total_committees} = {p_exactly_2_and_2:.4f}")

print()

# ---------------------------------------------------------------
# PROBLEM 3: Independent Repeated Trials — Coin Flipped 5 Times
# ---------------------------------------------------------------
# Q: What's P(getting exactly 3 Heads in 5 fair coin flips)?
# This is a Binomial scenario: n=5 trials, k=3 successes, p=0.5

n_flips = 5
k_heads = 3
p_head = 0.5
p_tail = 1 - p_head

# Ways to arrange 3 heads in 5 flips: 5C3
arrangements = combinations(n_flips, k_heads)   # = 10
p_exactly_3_heads = arrangements * (p_head ** k_heads) * (p_tail ** (n_flips - k_heads))

print("=== PROBLEM 3: Exactly 3 Heads in 5 Flips ===")
print(f"Arrangements of 3H in 5 flips: 5C3 = {arrangements}")
print(f"P(H)^3 = {p_head}^3 = {p_head**k_heads}")
print(f"P(T)^2 = {p_tail}^2 = {p_tail**(n_flips-k_heads)}")
print(f"P(exactly 3 Heads) = {arrangements} × {p_head**k_heads} × {p_tail**2} = {p_exactly_3_heads:.4f}")