To find the probability of obtaining more than three 2's when rolling 5 dice, we can use the binomial distribution. Here's a detailed step-by-step solution:
1. Define the Problem:
- We have 5 dice rolls (each roll is an independent trial), so the number of trials, [tex]\( n \)[/tex], is 5.
- The probability of rolling a 2 on any single die is [tex]\( \frac{1}{6} \)[/tex].
2. Understand the Binomial Distribution:
- In a binomial distribution, the probability of getting exactly [tex]\( k \)[/tex] successes (in this case, rolling a 2) in [tex]\( n \)[/tex] trials is given by the binomial probability formula:
[tex]\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}
\][/tex]
where [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient.
3. Calculate the Complement:
- Instead of directly calculating the probability of getting more than three 2's, we can use the cumulative distribution function (CDF) to find the probability of getting 0, 1, 2, or 3 2's, and subtract this from 1.
- This is because:
[tex]\[
P(X > 3) = 1 - P(X \leq 3)
\][/tex]
4. Use the Binomial CDF:
- To find [tex]\( P(X \leq 3) \)[/tex] using a graphing calculator, we use the binomial CDF function:
[tex]\[
\operatorname{binomcdf}\left(n, p, k\right)
\][/tex]
- Plugging in the values:
[tex]\[
\operatorname{binomcdf}\left(5, \frac{1}{6}, 3\right)
\][/tex]
5. Subtract from 1:
- To find the probability of more than three 2's, we subtract the cumulative probability from 1:
[tex]\[
P(X > 3) = 1 - \operatorname{binomcdf}\left(5, \frac{1}{6}, 3\right)
\][/tex]
6. Solution on a Graphing Calculator:
- Enter the values into the graphing calculator using the CDF function.
- Find [tex]\(\operatorname{binomcdf}(5, \frac{1}{6}, 3)\)[/tex].
- Subtract this value from 1.
Using this method, we find that the probability of obtaining more than three 2's when rolling 5 dice is approximately:
[tex]\[ P(X > 3) \approx 0.00334 \][/tex]
Therefore, the correct choice from the given options for this calculation would be:
[tex]\[ 1-\operatorname{binomcdf}\left(5, \frac{1}{6}, 3\right) \][/tex]
This method gives us the precise probability we are looking for, which is approximately 0.00334.
