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If a fair die is rolled 6 times, what is the probability, to the nearest thousandth, of getting exactly 4 sixes?
Answer:

If a fair die is rolled $6$ times, what is the probability, to the nearest thousandth, of getting exactly $4$ sixes? $\newline$ Answer:

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Probability of independent and dependent events

Full solution

Q. If a fair die is rolled

6

times, what is the probability, to the nearest thousandth, of getting exactly

4

sixes?

\newline

Answer:

Identify type of probability problem: Identify the type of probability problem. $\newline$ We are dealing with a binomial probability problem because we have a fixed number of independent trials (rolling a die $6$ times), two possible outcomes (rolling a six or not rolling a six), and we want to find the probability of getting a specific number of successes (rolling exactly $4$ sixes).
Calculate probability of success: Calculate the probability of success on a single trial. $\newline$ The probability of rolling a six on a fair die is $\frac{1}{6}$ , since there are $6$ sides and only one of them is a six.
Calculate probability of failure: Calculate the probability of failure on a single trial. The probability of not rolling a six (failure) is $\frac{5}{6}$ , since there are $5$ sides that are not a six.
Determine number of ways: Determine the number of ways to choose $4$ successes in $6$ trials. $\newline$ We use the combination formula to find the number of ways to choose $4$ successes (rolling a six) out of $6$ trials, which is denoted as ＂ $6$ choose $4$ ＂ or $C(6, 4)$ . $\newline$ $C(6, 4) = \frac{6!}{4! * (6-4)!} = \frac{(6 * 5 * 4 * 3 * 2 * 1)}{((4 * 3 * 2 * 1) * (2 * 1))} = \frac{(6 * 5)}{(2 * 1)} = 15$
Use binomial probability formula: Use the binomial probability formula to calculate the probability of exactly $4$ sixes in $6$ rolls. $\newline$ The binomial probability formula is $P(X = k) = C(n, k) \cdot (p^k) \cdot ((1-p)^{(n-k)})$ , where $P(X = k)$ is the probability of $k$ successes in $n$ trials, $C(n, k)$ is the number of combinations, $p$ is the probability of success, and $(1-p)$ is the probability of failure. $\newline$ $P(X = 4) = C(6, 4) \cdot (\frac{1}{6})^4 \cdot (\frac{5}{6})^{(6-4)}$ $\newline$ $P(X = 4) = 15 \cdot (\frac{1}{6})^4 \cdot (\frac{5}{6})^2$ $\newline$ $P(X = 4) = 15 \cdot (\frac{1}{1296}) \cdot (\frac{25}{36})$ $\newline$ $P(X = k)$ $0$ $\newline$ $P(X = k)$ $1$ $\newline$ $P(X = k)$ $2$ (rounded to the nearest thousandth)

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