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If a fair die is rolled 5 times, what is the probability, rounded to the nearest thousandth, of getting at most 2 threes? Answer:

If a fair die is rolled 55 times, what is the probability, rounded to the nearest thousandth, of getting at most 22 threes?\newlineAnswer:

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Q. If a fair die is rolled 55 times, what is the probability, rounded to the nearest thousandth, of getting at most 22 threes?\newlineAnswer:
  1. Understand the problem: Understand the problem.\newlineWe need to calculate the probability of rolling at most 22 threes in 55 rolls of a fair die. This means we need to find the probability of rolling 00, 11, or 22 threes.
  2. Calculate probabilities: Calculate the probability of rolling exactly 00, 11, or 22 threes.\newlineThe probability of rolling a three on a single roll is 16\frac{1}{6}, and the probability of not rolling a three is 56\frac{5}{6}.
  3. Calculate 00 threes: Calculate the probability of rolling exactly 00 threes.\newlineThis is the same as rolling a non-three 55 times in a row, which is (56)5(\frac{5}{6})^5.
  4. Calculate 11 three: Calculate the probability of rolling exactly 11 three.\newlineThis can happen in 55 different ways (one for each roll being the three), and the probability for each way is (16)×(56)4(\frac{1}{6}) \times (\frac{5}{6})^4. So, the total probability is 5×(16)×(56)45 \times (\frac{1}{6}) \times (\frac{5}{6})^4.
  5. Calculate 22 threes: Calculate the probability of rolling exactly 22 threes.\newlineThis can happen in several different ways. We can use the binomial coefficient to determine the number of ways to choose 22 rolls out of 55 to be threes. The binomial coefficient is ＂55 choose 22＂, which is 5!/(2!∗(5−2)!)5! / (2! * (5-2)!). This equals 1010. The probability for each way is (1/6)2∗(5/6)3(1/6)^2 * (5/6)^3. So, the total probability is 10∗(1/6)2∗(5/6)310 * (1/6)^2 * (5/6)^3.
  6. Add probabilities: Add the probabilities of rolling exactly 00, 11, or 22 threes to get the total probability.\newlineP(0 threes)+P(1 three)+P(2 threes)=(56)5+5×(16)×(56)4+10×(16)2×(56)3P(0 \text{ threes}) + P(1 \text{ three}) + P(2 \text{ threes}) = (\frac{5}{6})^5 + 5 \times (\frac{1}{6}) \times (\frac{5}{6})^4 + 10 \times (\frac{1}{6})^2 \times (\frac{5}{6})^3.
  7. Calculate and add: Calculate the probabilities and add them together.\newlineP(0 threes)=(56)5=0.40188P(0 \text{ threes}) = (\frac{5}{6})^5 = 0.40188 approximately,\newlineP(1 three)=5×(16)×(56)4=0.40188P(1 \text{ three}) = 5 \times (\frac{1}{6}) \times (\frac{5}{6})^4 = 0.40188 approximately,\newlineP(2 threes)=10×(16)2×(56)3=0.16075P(2 \text{ threes}) = 10 \times (\frac{1}{6})^2 \times (\frac{5}{6})^3 = 0.16075 approximately.\newlineAdding them together gives 0.40188+0.40188+0.16075=0.964510.40188 + 0.40188 + 0.16075 = 0.96451.
  8. Round the result: Round the result to the nearest thousandth.\newlineThe total probability rounded to the nearest thousandth is 0.9650.965.

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