To find the probability of obtaining at most 30 fives in 200 tosses of a fair die, we can use the binomial probability formula. However, since the number of tosses is large and the expected number of fives will determine the distribution closely resembles a normal distribution, we can apply the normal approximation for simplicity.
Overview of the Problem
A fair die has six faces, so the probability of rolling a five on any single toss is:
P(five) = 1/6
Thus, the probability of not rolling a five is:
P(not five) = 5/6
Parameters for Binomial Distribution
In this scenario, we can define:
- n = 200 (the number of tosses)
- p = 1/6 (the probability of rolling a five)
- q = 5/6 (the probability of not rolling a five)
- x = 30 (the maximum number of fives we are interested in)
Mean and Standard Deviation
The mean (μ) and standard deviation (σ) of the binomial distribution can be calculated as follows:
- Mean (μ): μ = n * p = 200 * (1/6) = 33.33
- Standard Deviation (σ): σ = √(n * p * q) = √(200 * (1/6) * (5/6)) = √(200 * 0.1667 * 0.8333) ≈ 4.08
Using Normal Approximation
Now that we have the mean and standard deviation, we can use the normal distribution to approximate the probability. We standardize our x value (30) to find the corresponding z-score using the formula:
z = (x – μ) / σ = (30 – 33.33) / 4.08 ≈ -0.81
Finding the Probability
Using the z-score, we consult the standard normal distribution table or use a calculator that provides cumulative probabilities. A z-score of -0.81 corresponds to a probability of approximately 0.2101.
This means that the probability of rolling at most 30 fives in 200 tosses of a die is about:
P(X ≤ 30) ≈ 21.01%
Conclusion
Therefore, the probability of obtaining at most 30 fives in 200 tosses of a fair die is approximately 21.01%. This calculation shows that while it is possible to achieve this outcome, it is slightly less than a quarter of the total cases, emphasizing the rarity of obtaining so few fives when tossing a die 200 times.