A 6-sided die is rolled 14 times. What is the probability of seeing a "6" at least 2 times?
I am sooo clueless please help!
Since you want at least two sixes, figure the probability of 0 or 1 six, then take the complement:
P(0) = (5/6)^14 * (1/6)^0 * 14!/0!/14!
P(1) = (5/6)^13 * (1/6)^1 * 14!/1!13!
calculate that and subtract it from 1. That's your answer.
To find the probability of seeing a "6" at least 2 times when rolling a 6-sided die 14 times, you can use the concept of binomial probability.
The probability of seeing a "6" on a single roll of a fair 6-sided die is 1/6, as there is one favorable outcome (rolling a "6") out of six possible outcomes (rolling a 1, 2, 3, 4, 5, or 6).
We can break down the problem into three separate cases:
1. Seeing exactly 2 "6"s.
2. Seeing exactly 3 "6"s or more.
3. Seeing exactly 4 "6"s or more.
For case 1:
The probability of seeing exactly 2 "6"s can be calculated using the binomial probability formula:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes,
- n is the number of trials (14 rolls of the die in this case),
- k is the number of desired successes (2 in this case),
- p is the probability of success on a single trial (1/6 in this case).
Using the formula, we can calculate the probability of getting exactly 2 "6"s:
P(X = 2) = (14 C 2) * (1/6)^2 * (5/6)^(14 - 2)
For case 2:
The probability of seeing exactly 3 "6"s or more can be calculated by summing up the probabilities of each individual outcome:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) + ... + P(X = 14)
You can use the same formula as before to calculate each individual probability and then sum them up.
For case 3:
The probability of seeing exactly 4 "6"s or more can also be calculated by summing up the probabilities of each individual outcome:
P(X ≥ 4) = P(X = 4) + P(X = 5) + ... + P(X = 14)
Again, you can use the same formula as before to calculate each individual probability and then sum them up.
Once you have calculated the probabilities for these three cases, you can add them together to get the total probability of seeing a "6" at least 2 times:
P(2 or more 6's) = P(X = 2) + P(X ≥ 3) + P(X ≥ 4)