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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 64%. What is the probability that it will rain on at most two of the six 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 64% 64 \% . What is the probability that it will rain on at most two of the six 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 64% 64 \% . What is the probability that it will rain on at most two of the six days they are there? Round your answer to the nearest thousandth.\newlineAnswer:
  1. Identify probability and days: Identify the probability of rain on any given day and the total number of days.\newlineProbability of rain on any given day: 64%64\% or 0.640.64\newlineTotal number of days: 66\newlineWe need to calculate the probability of it raining on at most two of those days.
  2. Calculate no rain probability: Calculate the probability of it not raining on any given day.\newlineProbability of no rain on any given day = 1−1 - Probability of rain on any given day\newline= 1−0.641 - 0.64\newline= 0.360.36
  3. Use binomial probability formula: Use the binomial probability formula to calculate the probability of it raining on exactly 00, 11, and 22 days.\newlineThe binomial probability formula 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:\newline- P(X=k)P(X = k) is the probability of kk successes in nn trials,\newline- (nk)\binom{n}{k} is the binomial coefficient,\newline- pp is the probability of success on a single trial,\newline- (1−p)(1-p) is the probability of failure on a single trial.\newlineWe will calculate this for 1100, 11, and 22.
  4. Calculate probability for 00 days: Calculate the probability of it raining on exactly 00 days k=0k=0.P(X=0)=(60)×(0.640)×(0.366)P(X = 0) = \binom{6}{0} \times (0.64^0) \times (0.36^6)=1×1×0.366= 1 \times 1 \times 0.36^6=0.366= 0.36^6
  5. Calculate probability for 11 day: Calculate the probability of it raining on exactly 11 day k=1k=1.P(X=1)=(61)×(0.641)×(0.365)P(X = 1) = \binom{6}{1} \times (0.64^1) \times (0.36^5) = 6×0.64×0.3656 \times 0.64 \times 0.36^5
  6. Calculate probability for 22 days: Calculate the probability of it raining on exactly 22 days k=2k=2.P(X=2)=(62)×(0.642)×(0.364)P(X = 2) = \binom{6}{2} \times (0.64^2) \times (0.36^4)=15×0.642×0.364= 15 \times 0.64^2 \times 0.36^4
  7. Add probabilities for total: Add the probabilities from steps 44, 55, and 66 to find the total probability of it raining on at most 22 days.\newlineTotal probability = P(X=0)+P(X=1)+P(X=2)P(X = 0) + P(X = 1) + P(X = 2)\newline= 0.366+(6×0.64×0.365)+(15×0.642×0.364)0.36^6 + (6 \times 0.64 \times 0.36^5) + (15 \times 0.64^2 \times 0.36^4)
  8. Perform calculations and round: Perform the calculations and round the answer to the nearest thousandth.\newlineTotal probability ≈0.366+(6×0.64×0.365)+(15×0.642×0.364)\approx 0.36^6 + (6 \times 0.64 \times 0.36^5) + (15 \times 0.64^2 \times 0.36^4)\newline≈0.00217678+(6×0.64×0.0089541)+(15×0.4096×0.01679616)\approx 0.00217678 + (6 \times 0.64 \times 0.0089541) + (15 \times 0.4096 \times 0.01679616)\newline≈0.00217678+0.03444288+0.10374144\approx 0.00217678 + 0.03444288 + 0.10374144\newline≈0.1403611\approx 0.1403611\newlineRounded to the nearest thousandth: 0.1400.140

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