If a fair die is rolled 4 times, what is the probability, to the nearest thousandth, of getting exactly 2 ones?
If a fair die is rolled 4 times, what is the probability, to the nearest thousandth, of getting exactly 2 ones?
Answer:
To find the probability of getting exactly 2 ones when a fair die is rolled 4 times, we can use the binomial probability formula. The binomial probability formula is:
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
Where:
- P(X = k) is the probability of getting exactly k successes in n trials.
- \binom{n}{k} is the binomial coefficient, which represents the number of ways to choose k successes out of n trials.
- p is the probability of success on a single trial.
- 1 - p is the probability of failure on a single trial.
- n is the total number of trials.
- k is the number of successes we are interested in.
For this problem:
- n = 4 (since the die is rolled 4 times)
- k = 2 (we want exactly 2 ones)
- p = \frac{1}{6} (the probability of rolling a one on a single die roll)
- 1 - p = \frac{5}{6} (the probability of not rolling a one on a single die roll)
The binomial coefficient \binom{n}{k} can be calculated as follows:
\binom{4}{2} = \frac{4!}{2!(4 - 2)!} = \frac{4!}{2! \cdot 2!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6
Next, we need to calculate p^k and (1 - p)^{n - k}:
p^k = \left(\frac{1}{6}\right)^2 = \frac{1}{36}
(1 - p)^{n - k} = \left(\frac{5}{6}\right)^2 = \frac{25}{36}
Now we can put it all together:
P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 = 6 \cdot \frac{1}{36} \cdot \frac{25}{36} = 6 \cdot \frac{25}{1296} = \frac{150}{1296} \approx 0.116
Therefore, the probability of getting exactly 2 ones when a fair die is rolled 4 times, to the nearest thousandth, is 0.116.