Math Problem Statement

When a fair dice is thrown four times by Mai, find the probability that she gets 6 with at least one of the four dice.

Solution

This problem involves calculating the probability of getting at least one "6" when a fair die is thrown four times. Let me solve it step by step:


Step 1: Probability of NOT getting a "6" in a single roll

For a fair die, the probability of getting any specific number (like "6") is 16\frac{1}{6}. Therefore, the probability of NOT getting a "6" is:

P(Not 6 in one roll)=116=56.P(\text{Not 6 in one roll}) = 1 - \frac{1}{6} = \frac{5}{6}.


Step 2: Probability of NOT getting a "6" in all four rolls

If the die is rolled four times, the events are independent. Thus, the probability of not getting a "6" in all four rolls is:

P(Not 6 in all 4 rolls)=(56)4.P(\text{Not 6 in all 4 rolls}) = \left( \frac{5}{6} \right)^4.

Let us compute this value: P(Not 6 in all 4 rolls)=(56)4=62512960.4823.P(\text{Not 6 in all 4 rolls}) = \left( \frac{5}{6} \right)^4 = \frac{625}{1296} \approx 0.4823.


Step 3: Probability of getting at least one "6"

The complement of "not getting a 6 in all rolls" is "getting at least one 6". Using the complement rule:

P(At least one 6)=1P(Not 6 in all 4 rolls).P(\text{At least one 6}) = 1 - P(\text{Not 6 in all 4 rolls}).

Substitute the value: P(At least one 6)=16251296=67112960.5177.P(\text{At least one 6}) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.5177.


Final Answer:

The probability that Mai gets at least one "6" in four rolls of the die is:

6711296 or approximately 51.77%.\boxed{\frac{671}{1296} \text{ or approximately } 51.77\%.}


Would you like further clarification or a deeper breakdown?

Here are 5 related questions for practice:

  1. What is the probability of getting exactly one "6" in four rolls of a fair die?
  2. What is the probability of getting no "6"s when the die is rolled six times?
  3. If a die is rolled three times, what is the probability of getting at least one even number?
  4. How does the probability of getting at least one "6" change if the die is rolled five times instead of four?
  5. What is the probability of getting at least two "6"s in four rolls of a fair die?

Tip: Always look for ways to simplify problems by using complementary probabilities, as it often reduces calculations.

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Math Problem Analysis

Mathematical Concepts

Probability
Independent Events
Complementary Probability

Formulas

P(Not 6 in one roll) = 1 - P(6 in one roll)
P(Not 6 in all rolls) = (P(Not 6 in one roll))^n
P(At least one 6) = 1 - P(Not 6 in all rolls)

Theorems

Complement Rule in Probability

Suitable Grade Level

Grades 9-12