Problem 42 - A First Course in Probability

Question

Two dice are thrown \(n\) times in succession. Compute the probability that double 6 appears at least once. How large need \(n\) be to make this probability at least \(\frac{1}{2} ?\)

Step-by-Step Solution

Answer

The probability of getting at least one double 6 in n trials is \(1 - \left(\frac{35}{36}\right)^n\). The smallest value of n for which this probability is at least \(\frac{1}{2}\) is 24.

Step 1: Find the probability of not getting a double 6 in a single trial

We know that a single die has 6 faces, numbered from 1 to 6. When two dice are thrown, there are 6 x 6 = 36 possible outcomes. We are interested in the probability of not getting a double 6, which is (6, 6). There is only 1 double 6, thus there are 36 - 1 = 35 favorable outcomes for not getting a double 6. Then, the probability of not getting a double 6 in a single trial can be written as: \(P(\text{not } \text{double} \: 6\) (Single Trial)) = \(\frac{35}{36}\)

Step 2: Find the probability of not getting a double 6 in n trials

We assume that each throw is independent. Therefore, the probability of not getting a double 6 in n trials is given by: \(P(\text{not } \text{double} \: 6 \: \text{in} \: n \: \text{trials}) = P(\text{not } \text{double} \: 6 \: \text{(Single Trial)})^n = \left(\frac{35}{36}\right)^n\)

Step 3: Find the probability of getting at least one double 6 in n trials

Now, we use the complementary probability to find the probability of getting at least one double 6 in n trials: \(P(\text{at least one double } 6 \: \text{in} \: n \: \text{trials}) = 1 - P(\text{not } \text{double} \: 6 \: \text{in} \: n \: \text{trials})\) = \(1 - \left(\frac{35}{36}\right)^n\)

Step 4: Find the smallest value of n for which the probability is at least 1/2

We want to find the smallest n such that \(P(\text{at least one double } 6 \: \text{in} \: n \: \text{trials})\) \(\geq \frac{1}{2}\). We can substitute the expression for this probability from Step 3 and solve for n: \(1 - \left(\frac{35}{36}\right)^n \geq \frac{1}{2}\) \(\left(\frac{35}{36}\right)^n \leq \frac{1}{2}\) To solve for n, we can use the logarithm. We have: \(n \log{\left(\frac{35}{36}\right)} \leq \log{\left(\frac{1}{2}\right)}\) Divide by the negative value of the logarithm of \(\frac{35}{36}\): \(n \geq \frac{\log{\left(\frac{1}{2}\right)}}{\log{\left(\frac{35}{36}\right)}}\) Using a calculator, we find that the smallest integer value of n that satisfies this inequality is: n = 24 Thus, at least 24 trials are needed to make the probability of getting at least one double 6 greater than or equal to 1/2.