Question: If a fair six-sided die is rolled 4 times, what is the probability that exactly two rolls show a prime number and the sum of all four rolls is even?

["# Probability of Rolling Exactly Two Prime Numbers and an Even Sum with a Fair Die", "When rolling a fair six-sided die four times, players often ask: What is the probability that exactly two rolls show a prime number AND the total sum of all four rolls is even? This probabilistic question combines two independent requirements—counting specific numbers and controlling the parity of the sum—making it a compelling challenge in discrete probability. In this article, we break down the solution step-by-step, combining number theory, combinatorics, and probability to find the exact answer.", "## Understanding the Die Sides and Prime Numbers", "A standard die has six faces: {1, 2, 3, 4, 5, 6}. Among these, the prime numbers are 2, 3, and 5. So, rolling a prime occurs with probability:\n[\nP(\ ext{prime}) = \frac{3}{6} = \frac{1}{2}\n]\nRolling a non-prime (1, 4, 6) also has probability:\n[\nP(\ ext{non-prime}) = \frac{3}{6} = \frac{1}{2}\n]", "Thus, exactly two out of four rolls showing primes follows a binomial distribution with parameters ( n = 4 ), ( p = \frac{1}{2} ). The number of ways to choose 2 prime rolls out of 4 is:\n[\n\binom{4}{2} = 6\n]\nThe probability of any specific pattern with exactly two primes and two non-primes is:\n[\n\left(\frac{1}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\n]\nSo, the marginal probability of exactly two prime rolls is:\n[\n6 \cdot \frac{1}{16} = \frac{6}{16} = \frac{3}{8}\n]\nHowever, this alone doesn’t account for the even sum condition, which adds critical structure to the problem.", "## Analyzing the Sum Parity Condition", "The total sum of four rolls is even if and only if:\n- The number of odd results among the four rolls is even (i.e., 0, 2, or 4 odd values), because odd + odd = even, and even + even = even.", "On a die:\n- Odd faces: 1, 3, 5 → prime and odd; total 3 odd\n- Even faces: 2, 4, 6 → 2 prime (only 2), 1 non-prime (4, 6) → half prime, all even", "Thus, each roll independently contributes:\n- Odd number: with probability ( \frac{3}{6} = \frac{1}{2} ) (since 3 odd numbers)\n- Even number: with probability ( \frac{1}{2} )", "But importance lies in how primes and parities overlap:\n- Prime 2: even, and prime\n- 3: odd, prime\n- 5: odd, prime\n- 1, 4, 6: even, non-prime", "So we classify die faces by both primality and parity:", "| Number | Prime? | Odd? | Write as odd/even | Prime and odd? | Prime and even? |\n|--------|--------|------|--------------------|----------------|-----------------|\n| 1 | No | Yes | Odd | No | No |\n| 2 | Yes | No | Even | Yes | No |\n| 3 | Yes | Yes | Odd | Yes | No |\n| 4 | No | Yes | Even | No | Yes |\n| 5 | Yes | Yes | Odd | Yes | No |\n| 6 | No | No | Even | No | Yes |", "Define for each roll:\n- An "odd prime" face is 3 or 5 → two outcomes\n- An "even non-prime" is 4 or 6 → two outcomes\n- Total odd rolls: 1 (only 3) plus odd primes\n- Even rolls: 1 (2) plus even non-primes", "Let:\n- ( O ): odd total sum → number of odd faces ≡ 0 or 2 or 4 mod 2\nBut we want even total sum → number of odd faces in 4 rolls must be even: 0, 2, or 4.", "We now condition on the constraint: exactly two prime faces, and even sum (i.e., even number of odd faces).", "We proceed by enumerating valid outcomes where:\n- Exactly 2 prime faces (2 are primes: choices of 2 out of 3 primes: 2, 3, 5)\n- The remaining 2 faces are non-primes (1, 4, 6)\n- Total number of odd faces (from roll types) is even (0, 2, or 4)", "We analyze based on how many odd numbers appear among the four rolls, given the constraint of exactly two primes.", "### Step 1: Enumerate all prime/non-prime combinations with exactly 2 primes", "There are ( \binom{4}{2} = 6 ) positions for the two prime rolls. For each pair of positions, we assign primes and non-primes, then classify by parity.", "Let’s define:\n- A prime roll is either odd (3 or 5) or even (2)\n- A non-prime roll is odd (1 or 5? Wait: 5 is odd prime — correction: non-primes are 1, 4, 6 — only 1 is odd; 4, 6 even)", "So:\n- Odd primes: 3, 5 → 2 choices\n- Even primes: 2 → 1 choice\n- Odd non-prime: 1 → 1 choice\n- Even non-prime: 4, 6 → 2 choices", "We now fix the 2 prime rolls and 2 non-prime rolls, and compute for each case whether the total number of odd faces is even (to satisfy even sum), and sum probabilities accordingly.", "But due to symmetry in non-prime outcomes, we can compute expectations per case.", "Instead, use conditional counting with generating cases based on odd/even roll types.", "Let’s define:\nEach prime roll is either:\n- Odd: 3 or 5 → 2 outcomes\n- Even: 2 → 1 outcome\nEach non-prime roll is:\n- Odd: only 1 → 1 outcome\n- Even: 4 or 6 → 2 outcomes", "We consider all combinations of which two positions are prime (6 ways), but symmetry lets us compute per type group, not individual positions.", "We define types:", "Let’s define a case by:\n- Number of odd prime rolls among the two prime positions: can be 0, 1, or 2\n- Number of odd non-prime rolls among the two non-prime positions: 0, 1, or 2\nTotal odd faces = (2 × odd prime) contributes:\n- If odd prime (3 or 5): contributes 1 odd face\n- If even prime (2): contributes 0 odd face", "Similarly, odd non-prime (only 1): contributes 1 odd face", "So:", "Let:\n- ( k ): number of odd primes among 2 prime rolls → ( k = 0,1,2 )\n- ( m ): number of odd non-primes among 2 non-prime rolls → ( m = 0,1,2 )", "Total odd faces = ( k ) (from primes) + ( m ) (from non-primes)\nWe want this sum even:\n[\nk + m \equiv 0 \pmod{2}\n]", "Now, for each ( k ) and ( m ) such that ( k + m ) even, compute probability.", "We compute for each valid ( (k, m) ):", "#### Case 1: ( k = 0 ), ( m = 0 )\n- No odd primes (i.e., two even primes: 2 and 2)\n- No odd non-primes (i.e., two even non-primes: 4 and 6)\n- Total odd faces: 0 → even → valid\n- Number of ways:\n - Prime rolls: both 2 → 1 × 1 = 1 way\n - Non-prime rolls: both 4 or 6 → 2 × 2 = 4 ways\n - Total favorable prime/non-prime assignments: ( 1 \ imes 4 = 4 )\n - Position: number of ways to assign 2 primes to 4 rolls: ( \binom{4}{2} = 6 ) → all valid under this type\n - But for each such positional split, the specific face values matter.", "For each prime position assigned 2: 1 choice (only 2)\nFor each non-prime position assigned 4 or 6: 2 choices each → total 4 options", "So for fixed positions (say rolls 1,2 prime; 3,4 non-prime):\n- Prime faces: both 2 → 1 way\n- Non-prime faces: 4 or 6 → 2 × 2 = 4 ways → total 4 outcome combinations", "Total over all positional splits: still ( \binom{4}{2} = 6 ) ways to place prime rolls, but for each, probability depends only on types.", "Since all configurations with ( k=0, m=0 ) have:\n- Prime faces: both 2 → prob per roll: ( \frac{1}{6} \ imes \frac{1}{6} = \frac{1}{36} ), but conditional on being prime, but we must multiply by probability of face values.", "Better: compute total probability as sum over valid configurations.", "Each configuration (bit assignment: which are prime, and their values) has probability:\n[\n\left(\frac{1}{6}\right)^2 \ imes \left(\frac{1}{6}\right)^2 = \frac{1}{1296}\n]\nBut number of such configurations satisfying the conditions is what we sum.", "Total number of valid 4-tuples with exactly two primes and even sum:\nWe compute by enumerating ( (k,m) ) pairs with ( k + m ) even.", "Let’s tabulate:", "| k | m | k+m mod 2 | Valid? |\n|---|---|-----------|--------|\n| 0 | 0 | 0 | Yes |\n| 0 | 1 | 1 | No |\n| 0 | 2 | 2 | Yes |\n| 1 | 0 | 1 | No |\n| 1 | 1 | 2 | Yes |\n| 1 | 2 | 3 | No |\n| 2 | 0 | 2 | Yes |\n| 2 | 1 | 3 | No |\n| 2 | 2 | 4 | Yes |", "So valid: (0,0), (0,2), (1,1), (2,0), (2,2)", "Now compute number of favorable outcomes for each.", "---", "### Case 1: ( k = 0 ), ( m = 0 )\n- No odd primes: both prime rolls are 2 → only 1 way per prime roll → total prime combinations: ( 1 \ imes 1 = 1 )\n- No odd non-primes: both non-prime rolls are 4 or 6 → 2 choices each → ( 2 \ imes 2 = 4 )\n- Number of ways to assign 2 primes to 4 rolls: ( \binom{4}{2} = 6 )\n- For each such assignment, number of value combinations: ( 1 \ imes 4 = 4 )\n- Total outcomes: ( 6 \ imes 4 = 24 )\n- Each outcome has probability: ( \left(\frac{1}{6}\right)^4 = \frac{1}{1296} )\n- Total probability: ( 24 \ imes \frac{1}{1296} = \frac{24}{1296} )", "---", "### Case 2: ( k = 0 ), ( m = 2 )\n- No odd primes → both primes are 2 → 1 way\n- No odd non-primes → both non-primes are 4 or 6 → 4 ways\n- Same as above: 6 positional splits, 4 value combos per, 24 total value combinations\n- Probability: ( \frac{24}{1296} ) → same as case 1", "But note: in this case, odd count = 0 (primes) + 2 (from non-primes) = 2 → even → valid", "---", "### Case 3: ( k = 1 ), ( m = 1 )\n- One odd prime, one even prime\n- One odd non-prime, one even non-prime", "Number of ways to assign types:\n- Choose 1 of 2 prime positions for odd prime (3 or 5): ( \binom{2}{1} = 2 )\n- Remaining prime position: must be 2 (even prime) → 1 way\n- Among non-prime positions: one must be 1 (only odd), one must be 4 or 6\n - Assign 1 odd non-prime to one position: 1 choice (must pick 1)\n - Other to 4 or 6: 2 choices\n - But which position gets 1? 2 choices among non-"]