Question

In four throws, with a pair of dice, what is the probability of occurrence of doublets at least twice?

Hint:

For binomial problems, use the binomial distribution formula \( P(k) = \binom{n}{k} p^k q^{n-k} \) to calculate the probability for different outcomes.

Answer 1

A doublet occurs when both dice show the same number. The probability of getting a doublet in a single throw is: \[ P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}. \] The probability of not getting a doublet in a single throw is: \[ P(\text{not a doublet}) = 1 - \frac{1}{6} = \frac{5}{6}. \] We are asked to find the probability of getting at least two doublets in four throws. This is a binomial probability problem, where: - \( n = 4 \) (the number of trials), - \( p = \frac{1}{6} \) (the probability of success, i.e., getting a doublet), - \( q = \frac{5}{6} \) (the probability of failure, i.e., not getting a doublet). The probability of getting exactly \( k \) doublets in 4 throws is given by the binomial distribution: \[ P(k \text{ doublets}) = \binom{4}{k} p^k q^{4-k}. \] We need to calculate the probability of getting at least two doublets, i.e., \( P(k \ge 2) \), which can be expressed as: \[ P(k \ge 2) = 1 - P(k<2) = 1 - \left( P(k = 0) + P(k = 1) \right). \] Now, calculate \( P(k = 0) \) and \( P(k = 1) \): 1. For \( k = 0 \) (no doublets): \[ P(k = 0) = \binom{4}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^4 = 1 \cdot 1 \cdot \left(\frac{5}{6}\right)^4 = \left(\frac{5}{6}\right)^4 = \frac{625}{1296}. \] 2. For \( k = 1 \) (one doublet): \[ P(k = 1) = \binom{4}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^3 = 4 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^3 = 4 \cdot \frac{1}{6} \cdot \frac{125}{216} = \frac{500}{1296}. \] Now, calculate \( P(k \ge 2) \): \[ P(k \ge 2) = 1 - \left( \frac{625}{1296} + \frac{500}{1296} \right) = 1 - \frac{1125}{1296} = \frac{171}{1296}. \] Therefore, the probability of getting at least two doublets in four throws is: \[ \boxed{\frac{171}{1296}}. \]