Multiple choice

A cubical die is thrown $6$ times. The probability that $2$ and $4$ will turn up exactly $3$ times each is

  1. $\displaystyle \frac{1}{6^{6}}$
  2. $\displaystyle \frac{ 6!}{6^{6}}$
  3. $\displaystyle \frac{ 6!}{ 3! \; 3!}\times \displaystyle \frac{1}{6^{6}}$
  4. $1$
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C Correct answer
Explanation

This is a multinomial distribution problem. Probability of 2 is 1/6, 4 is 1/6, other is 4/6. We want 3 of 2, 3 of 4, 0 of other in 6 trials. Formula: (6! / (3!3!0!)) * (1/6)^3 * (1/6)^3 * (4/6)^0 = (6! / (3!3!)) * (1/6)^6.

AI explanation

The probability of rolling a 2 or a 4 is 1/6 for each individual outcome. We need exactly three 2s and exactly three 4s in six rolls, so the probability of this specific sequence is (1/6)^3 multiplied by (1/6)^3, which equals (1/6)^6. The number of ways to arrange these six outcomes is given by the multinomial coefficient 6! / (3! * 3!). Multiplying the arrangements by the sequence probability gives (6! / (3! * 3!)) * (1/6)^6.

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