If a fair die is rolled $5$ times, what is the probability, rounded to the nearest thousandth, of getting at least $4$ threes? $\newline$ 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 least

4

threes?

\newline

Answer:

Identify Cases: To solve this problem, we need to consider the cases where we get exactly $4$ threes and the case where we get exactly $5$ threes. 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 successes, ＂ $p$ ＂ is the probability of success on a single trial, and ＂ $\binom{n}{k}$ ＂ is the binomial coefficient.
Calculate Probability for $4$ Threes: First, let's calculate the probability of getting exactly $4$ threes. In this case, $n=5$ , $k=4$ , and $p=\frac{1}{6}$ (since there is one three on a six-sided die). The binomial coefficient $\binom{5}{4}$ is $5$ , because there are $5$ ways to get $4$ threes in $5$ rolls. So, $P(X=4) = \binom{5}{4} \times \left(\frac{1}{6}\right)^4 \times \left(\frac{5}{6}\right)^{5-4}$ .
Calculate Probability for $5$ Threes: Calculating $P(X=4)$ gives us $P(X=4) = 5 \times (\frac{1}{6})^4 \times (\frac{5}{6})^1 = 5 \times (\frac{1}{1296}) \times (\frac{5}{6}) = \frac{5}{1296} \times \frac{5}{6} = \frac{25}{7776}$ .
Add Probabilities for At Least $4$ Threes: Next, we calculate the probability of getting exactly $5$ threes. In this case, $n=5$ , $k=5$ , and $p=\frac{1}{6}$ . The binomial coefficient $(5 \choose 5)$ is $1$ , because there is only one way to get $5$ threes in $5$ rolls. So, $P(X=5) = (5 \choose 5) \times \left(\frac{1}{6}\right)^5 \times \left(\frac{5}{6}\right)^{5-5}$ .
Simplify Fraction: Calculating $P(X=5)$ gives us $P(X=5) = 1 \times (\frac{1}{6})^5 \times (\frac{5}{6})^0 = \frac{1}{7776}.$
Round to Nearest Thousandth: Now, we add the probabilities of getting exactly $4$ threes and exactly $5$ threes to find the total probability of getting at least $4$ threes. So, the total probability is $P(X\geq4) = P(X=4) + P(X=5) = \frac{25}{7776} + \frac{1}{7776}.$
Round to Nearest Thousandth: Now, we add the probabilities of getting exactly $4$ threes and exactly $5$ threes to find the total probability of getting at least $4$ threes. So, the total probability is $P(X\geq4) = P(X=4) + P(X=5) = \frac{25}{7776} + \frac{1}{7776}$ . Adding the probabilities gives us $P(X\geq4) = \frac{25}{7776} + \frac{1}{7776} = \frac{26}{7776}$ . This fraction can be simplified by dividing both the numerator and the denominator by $2$ , which gives us $\frac{13}{3888}$ .
Round to Nearest Thousandth: Now, we add the probabilities of getting exactly $4$ threes and exactly $5$ threes to find the total probability of getting at least $4$ threes. So, the total probability is $P(X\geq4) = P(X=4) + P(X=5) = \frac{25}{7776} + \frac{1}{7776}$ . Adding the probabilities gives us $P(X\geq4) = \frac{25}{7776} + \frac{1}{7776} = \frac{26}{7776}$ . This fraction can be simplified by dividing both the numerator and the denominator by $2$ , which gives us $\frac{13}{3888}$ . Finally, we round the probability to the nearest thousandth. The decimal form of $\frac{13}{3888}$ is approximately $0.00334$ , which rounds to $0.003$ .