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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 11%. What is the probability that it will rain on at most one 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 11% 11 \% . What is the probability that it will rain on at most one 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 11% 11 \% . What is the probability that it will rain on at most one of the six days they are there? Round your answer to the nearest thousandth.\newlineAnswer:
  1. Use Binomial Probability Formula: Use the binomial probability formula: P(X≤k)=Σ(C(n,i)⋅(p)i⋅(1−p)(n−i))P(X \leq k) = \Sigma (C(n, i) \cdot (p)^i \cdot (1-p)^{(n-i)}) for ii from 00 to kk. Identify the values of nn, kk, and pp. n=6n = 6 (the number of days they are there) k=1k = 1 (at most one day of rain) p=0.11p = 0.11 (the probability of rain on any given day)
  2. Identify Values of n,k,pn, k, p: Calculate the probability of no rain on any of the six days using the binomial probability formula.P(X=0)=C(6,0)⋅(p)0⋅(1−p)(6−0)P(X = 0) = C(6, 0) \cdot (p)^0 \cdot (1-p)^{(6-0)}Substitute n=6n = 6, k=0k = 0 and p=0.11p = 0.11 into the formula.P(X=0)=C(6,0)⋅(0.11)0⋅(1−0.11)6P(X = 0) = C(6, 0) \cdot (0.11)^0 \cdot (1 - 0.11)^{6}
  3. Calculate Probability of No Rain: Calculate the value of C(6,0)C(6, 0). \newlineC(6,0)=6!0!(6−0)!C(6, 0) = \frac{6!}{0! (6 - 0)!}\newlineC(6,0)=1C(6, 0) = 1 (since n!n!\frac{n!}{n!} is always 11 and 0!0! is 11)
  4. Calculate Value of C(6,0)C(6, 0): Solve (0.11)0(0.11)^0 and (1−0.11)6(1 - 0.11)^{6}.
    (0.11)0=1(0.11)^0 = 1 (any number to the power of 00 is 11)
    (1−0.11)6=(0.89)6(1 - 0.11)^{6} = (0.89)^6
    (0.89)6≈0.558405(0.89)^6 \approx 0.558405
  5. Solve (0.11)0(0.11)^0 and (1−0.11)6(1 - 0.11)^6: Calculate P(X=0)P(X = 0) using the values obtained.\newlineP(X=0)=1×1×0.558405P(X = 0) = 1 \times 1 \times 0.558405\newlineP(X=0)≈0.558405P(X = 0) \approx 0.558405
  6. Calculate P(X=0)P(X = 0): Calculate the probability of exactly one day of rain using the binomial probability formula.P(X=1)=C(6,1)⋅(p)1⋅(1−p)(6−1)P(X = 1) = C(6, 1) \cdot (p)^1 \cdot (1-p)^{(6-1)}Substitute n=6n = 6, k=1k = 1 and p=0.11p = 0.11 into the formula.P(X=1)=C(6,1)⋅(0.11)1⋅(1−0.11)5P(X = 1) = C(6, 1) \cdot (0.11)^1 \cdot (1 - 0.11)^{5}
  7. Calculate Probability of Exactly One Day of Rain: Calculate the value of C(6,1)C(6, 1). \newlineC(6,1)=6!1!(6−1)!C(6, 1) = \frac{6!}{1! (6 - 1)!}\newlineC(6,1)=61C(6, 1) = \frac{6}{1} (since 5!5! cancels out from the numerator and denominator)\newlineC(6,1)=6C(6, 1) = 6
  8. Calculate Value of C(6,1)C(6, 1): Solve (0.11)1(0.11)^1 and (1−0.11)(5)(1 - 0.11)^{(5)}.$0.11\$0.11^11 = 00.1111\)$1−0.11\$1 - 0.11^{(55)} = 0.890.89^55\)\(0.89^55 \approx 00.558405558405 \times 00.8989 \approx 00.4968004549680045\)
  9. Solve (0.11)1(0.11)^1 and (1−0.11)5(1 - 0.11)^5: Calculate P(X=1)P(X = 1) using the values obtained.\newlineP(X=1)=6×0.11×0.49680045P(X = 1) = 6 \times 0.11 \times 0.49680045\newlineP(X=1)≈0.328368297P(X = 1) \approx 0.328368297
  10. Calculate P(X=1)P(X = 1): Calculate the total probability of it raining on at most one day by adding P(X=0)P(X = 0) and P(X=1)P(X = 1).
    P(X≤1)=P(X=0)+P(X=1)P(X \leq 1) = P(X = 0) + P(X = 1)
    P(X≤1)≈0.558405+0.328368297P(X \leq 1) \approx 0.558405 + 0.328368297
    P(X≤1)≈0.886773297P(X \leq 1) \approx 0.886773297
  11. Calculate Total Probability: Round the answer to the nearest thousandth.\newlineP(X≤1)≈0.887P(X \leq 1) \approx 0.887

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