If a fair die is rolled $5$ times, what is the probability, rounded to the nearest thousandth, of getting at most $1$ four? $\newline$ Answer:

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Grade 8

Probability of independent and dependent events

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Q. If a fair die is rolled

5

times, what is the probability, rounded to the nearest thousandth, of getting at most

1

four?

\newline

Answer:

Determine Probability of Rolling: Determine the probability of rolling a four on a single die roll. $\newline$ A fair die has $6$ sides, so the probability of rolling any specific number, including a four, is $\frac{1}{6}$ . Conversely, the probability of not rolling a four is $\frac{5}{6}$ .
Calculate Probability of One Four: Calculate the probability of rolling exactly one four in five rolls. $\newline$ This can happen in several different ways: the four can appear on the first, second, third, fourth, or fifth roll. We will use the binomial probability formula, which is $P(X=k) = \binom{n}{k} \cdot (p^k) \cdot ((1-p)^{(n-k)})$ , where $n$ is the number of trials, $k$ is the number of successful outcomes we want, and $p$ is the probability of success on a single trial. $\newline$ For exactly one four, $k=1$ , $n=5$ , and $p=\frac{1}{6}$ . So we calculate: $\newline$ $P(X=1) = \binom{5}{1} \cdot (\frac{1}{6})^1 \cdot (\frac{5}{6})^{5-1}$ $\newline$ $P(X=1) = 5 \cdot (\frac{1}{6}) \cdot (\frac{5}{6})^4$
Calculate Probability of No Fours: Calculate the probability of rolling no fours in five rolls. $\newline$ For this scenario, $k=0$ , $n=5$ , and $p=\frac{1}{6}$ . We use the binomial probability formula again: $\newline$ $P(X=0) = \binom{5}{0} \times \left(\frac{1}{6}\right)^0 \times \left(\frac{5}{6}\right)^{5-0}$ $\newline$ $P(X=0) = 1 \times 1 \times \left(\frac{5}{6}\right)^5$
Add Probabilities for At Most One Four: Add the probabilities of rolling exactly one four and no fours to get the total probability of rolling at most one four. $\newline$ $P(\text{at most one } 4) = P(X=0) + P(X=1)$ $\newline$ $P(\text{at most one } 4) = (\frac{5}{6})^5 + 5 \cdot (\frac{1}{6}) \cdot (\frac{5}{6})^4$
Perform Calculations: Perform the calculations. $\newline$ $P(X=0) = \left(\frac{5}{6}\right)^5 = \left(\frac{5}{6}\right)^5 \approx 0.40188$ $\newline$ $P(X=1) = 5 \times \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right)^4 \approx 0.40188$ $\newline$ $P(\text{at most one 4}) = 0.40188 + 0.40188 \approx 0.80376$
Recalculate Probability of One Four: Step $5$ (Correction): Perform the correct calculations. $\newline$ $P(X=0) = (\frac{5}{6})^5 \approx 0.40188$ (corrected value) $\newline$ $P(X=1) = 5 \times (\frac{1}{6}) \times (\frac{5}{6})^4 \approx 0.40188$ (incorrect value, needs to be recalculated) $\newline$ $P(X=1) = 5 \times (\frac{1}{6}) \times (\frac{5}{6})^4 = 5 \times (\frac{1}{6}) \times (\frac{625}{1296}) \approx 0.32143$ $\newline$ $P(\text{at most one 4}) = P(X=0) + P(X=1) = 0.40188 + 0.32143 \approx 0.72331$
Round Final Probability: Round the final probability to the nearest thousandth. $\newline$ $P(\text{at most one } 4) \approx 0.723$ (rounded to the nearest thousandth)