Two Dice & At-Least-One Problems

🎯 Learning Objective

Handle 36-outcome two-dice questions and crack 'at least one' using the complement.

💡 Concept

• Two dice → 6 × 6 = 36 equally likely ordered pairs
• Sum counts: 2→1, 3→2, 4→3, 5→4, 6→5, 7→6, then mirror down to 12→1
• P(at least one) = 1 − P(none) — the golden shortcut
• Independent events: P(A and B) = P(A) × P(B)

🧮 Key Formulas

Two dice: total = 36

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P(at least one) = 1 − P(none)

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Independent: P(A and B) = P(A) × P(B)

✏️ Easy Example

Q. Two dice are thrown. Find the probability that the sum is 7.

1. Pairs: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)
2. Favourable = 6, total = 36
3. P = 6/36

Answer: 1/6

🇮🇳 Real-Life Example

Waiting for a six to open your token in Ludo: one throw gives 1/6, and in two throws P(at least one six) = 1 − (5/6)² = 11/36.

📝 Exam-Level Example

Q. Two coins are tossed together. Find the probability of getting at least one head.

1. P(no head) = P(TT) = 1/4
2. P(at least one head) = 1 − 1/4

Answer: 3/4

📝 Exam-Level Example

Q. Two dice are thrown. Find the probability that the sum is at least 10.

1. Sum 10: 3 ways, 11: 2 ways, 12: 1 way
2. Favourable = 6
3. P = 6/36

Answer: 1/6

🪄 Memory Trick

Two-dice sum s: favourable ways = s − 1 when s ≤ 7, and 13 − s when s > 7. Sum 9 → 13 − 9 = 4 ways.

⚠️ Common Mistakes

• ❌ Treating (2,5) and (5,2) as one outcome — the dice are distinct
• ❌ Computing 'at least one' by adding cases and double-counting
• ❌ Using 12 as the total number of two-dice outcomes instead of 36

🏆 Exam Tips

• ✅ 'At least one' → ALWAYS go via the complement
• ✅ Memorise: sum 7 has 6 ways — the most likely two-dice sum

📌 Summary

• Two dice = 36 ordered outcomes
• Sum-ways: s − 1 (s ≤ 7), 13 − s (s > 7)
• At least one = 1 − none
• Independent events multiply