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The Hiking Club plans to go camping in a State park where the probability of rain on any given day is 5//6. What is the probability that it will rain on at least five of the seven days they are there? Round your answer to the nearest thousandth. Answer:

The Hiking Club plans to go camping in a State park where the probability of rain on any given day is 5/6 5 / 6 . What is the probability that it will rain on at least five of the seven days they are there? Round your answer to the nearest thousandth.\newlineAnswer:

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Q. The Hiking Club plans to go camping in a State park where the probability of rain on any given day is 5/6 5 / 6 . What is the probability that it will rain on at least five of the seven days they are there? Round your answer to the nearest thousandth.\newlineAnswer:
  1. Understand the Problem: Understand the problem and determine the approach.\newlineWe need to calculate the probability of it raining on at least five out of seven days. Since the probability of rain on any given day is 56\frac{5}{6}, we can use the binomial probability formula to calculate the probability of it raining on exactly kk days out of nn days, which is P(X=k)=(nk)⋅(pk)⋅((1−p)(n−k))P(X = k) = \binom{n}{k} \cdot (p^k) \cdot ((1-p)^{(n-k)}), where pp is the probability of rain on a single day, and (nk)\binom{n}{k} is the binomial coefficient.
  2. Calculate Probability of 55 Days: Calculate the probability of it raining on exactly five days. Using the binomial probability formula, we calculate P(X=5)P(X = 5) where n=7n = 7, k=5k = 5, and p=56p = \frac{5}{6}. P(X=5)=(75)⋅(56)5⋅(16)2P(X = 5) = \binom{7}{5} \cdot \left(\frac{5}{6}\right)^5 \cdot \left(\frac{1}{6}\right)^2 P(X=5)=21⋅(31257776)⋅(136)P(X = 5) = 21 \cdot \left(\frac{3125}{7776}\right) \cdot \left(\frac{1}{36}\right) P(X=5)=21⋅(3125279936)P(X = 5) = 21 \cdot \left(\frac{3125}{279936}\right) P(X=5)=65625279936P(X = 5) = \frac{65625}{279936}
  3. Calculate Probability of 66 Days: Calculate the probability of it raining on exactly six days. Using the binomial probability formula, we calculate P(X=6)P(X = 6) where n=7n = 7, k=6k = 6, and p=56p = \frac{5}{6}. P(X=6)=(76)⋅(56)6⋅(16)1P(X = 6) = \binom{7}{6} \cdot \left(\frac{5}{6}\right)^6 \cdot \left(\frac{1}{6}\right)^1 P(X=6)=7⋅(1562546656)⋅(16)P(X = 6) = 7 \cdot \left(\frac{15625}{46656}\right) \cdot \left(\frac{1}{6}\right) P(X=6)=7⋅(15625279936)P(X = 6) = 7 \cdot \left(\frac{15625}{279936}\right) P(X=6)=109375279936P(X = 6) = \frac{109375}{279936}
  4. Calculate Probability of 77 Days: Calculate the probability of it raining on all seven days.\newlineUsing the binomial probability formula, we calculate P(X=7)P(X = 7) where n=7n = 7, k=7k = 7, and p=56p = \frac{5}{6}.\newlineP(X=7)=(77)⋅(56)7⋅(16)0P(X = 7) = \binom{7}{7} \cdot \left(\frac{5}{6}\right)^7 \cdot \left(\frac{1}{6}\right)^0\newlineP(X=7)=1⋅(78125279936)P(X = 7) = 1 \cdot \left(\frac{78125}{279936}\right)\newlineP(X=7)=78125279936P(X = 7) = \frac{78125}{279936}
  5. Add Probabilities for At Least 55 Days: Add the probabilities of it raining on exactly five, six, and seven days to find the total probability of it raining on at least five days.\newlineP(at least 55 days) = P(X=5)+P(X=6)+P(X=7)P(X = 5) + P(X = 6) + P(X = 7)\newlineP(at least 55 days) = 65625279936+109375279936+78125279936\frac{65625}{279936} + \frac{109375}{279936} + \frac{78125}{279936}\newlineP(at least 55 days) = (65625+109375+78125)279936\frac{(65625 + 109375 + 78125)}{279936}\newlineP(at least 55 days) = 253125279936\frac{253125}{279936}
  6. Round the Answer: Round the answer to the nearest thousandth.\newlineP(at least 5 days)≈253125279936≈0.904P(\text{at least } 5 \text{ days}) \approx \frac{253125}{279936} \approx 0.904 (rounded to three decimal places)

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