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. $\newline$ Answer:

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Find probabilities using the binomial distribution

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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 \%

. 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.

\newline

Answer:

Use Binomial Probability Formula: Use the binomial probability formula: $P(X \leq k) = \Sigma (C(n, i) \cdot (p)^i \cdot (1-p)^{(n-i)})$ for $i$ from $0$ to $k$ . Identify the values of $n$ , $k$ , and $p$ . $n = 6$ (the number of days they are there) $k = 1$ (at most one day of rain) $p = 0.11$ (the probability of rain on any given day)
Identify Values of $n, k, p$ : Calculate the probability of no rain on any of the six days using the binomial probability formula. $P(X = 0) = C(6, 0) \cdot (p)^0 \cdot (1-p)^{(6-0)}$ Substitute $n = 6$ , $k = 0$ and $p = 0.11$ into the formula. $P(X = 0) = C(6, 0) \cdot (0.11)^0 \cdot (1 - 0.11)^{6}$
Calculate Probability of No Rain: Calculate the value of $C(6, 0)$ . $\newline$ $C(6, 0) = \frac{6!}{0! (6 - 0)!}$ $\newline$ $C(6, 0) = 1$ (since $\frac{n!}{n!}$ is always $1$ and $0!$ is $1$ )
Calculate Value of $C(6, 0)$ : Solve $(0.11)^0$ and $(1 - 0.11)^{6}$ .
$(0.11)^0 = 1$ (any number to the power of $0$ is $1$ )
$(1 - 0.11)^{6} = (0.89)^6$
$(0.89)^6 \approx 0.558405$
Solve $(0.11)^0$ and $(1 - 0.11)^6$ : Calculate $P(X = 0)$ using the values obtained. $\newline$ $P(X = 0) = 1 \times 1 \times 0.558405$ $\newline$ $P(X = 0) \approx 0.558405$
Calculate $P(X = 0)$ : Calculate the probability of exactly one day of rain using the binomial probability formula. $P(X = 1) = C(6, 1) \cdot (p)^1 \cdot (1-p)^{(6-1)}$ Substitute $n = 6$ , $k = 1$ and $p = 0.11$ into the formula. $P(X = 1) = C(6, 1) \cdot (0.11)^1 \cdot (1 - 0.11)^{5}$
Calculate Probability of Exactly One Day of Rain: Calculate the value of $C(6, 1)$ . $\newline$ $C(6, 1) = \frac{6!}{1! (6 - 1)!}$ $\newline$ $C(6, 1) = \frac{6}{1}$ (since $5!$ cancels out from the numerator and denominator) $\newline$ $C(6, 1) = 6$
Calculate Value of $C(6, 1)$ : Solve $(0.11)^1$ and $(1 - 0.11)^{(5)}$ . $\$0.11$ ^ $1$ = $0$ . $11$ \) $\$1 - 0.11$ ^{( $5$ )} = $0.89$ ^ $5$ \)$0.89^ $5$ \approx $0$ . $558405$ \times $0$ . $89$ \approx $0$ . $49680045$ $
Solve $(0.11)^1$ and $(1 - 0.11)^5$ : Calculate $P(X = 1)$ using the values obtained. $\newline$ $P(X = 1) = 6 \times 0.11 \times 0.49680045$ $\newline$ $P(X = 1) \approx 0.328368297$
Calculate $P(X = 1)$ : Calculate the total probability of it raining on at most one day by adding $P(X = 0)$ and $P(X = 1)$ .
$P(X \leq 1) = P(X = 0) + P(X = 1)$
$P(X \leq 1) \approx 0.558405 + 0.328368297$
$P(X \leq 1) \approx 0.886773297$
Calculate Total Probability: Round the answer to the nearest thousandth. $\newline$ $P(X \leq 1) \approx 0.887$