Question

if a fair die is rolled 5 times what is the probability round to the nearest thousandth of getting atmost 1 five

Answer 1

Using the binomial probability formula, the probability of obtaining at most 1 "one" in 5 rolls of a fair six-sided die is approximately 0.672, rounded to the nearest thousandth.

To find the probability of getting at most 1 "one" when rolling a fair six-sided die 5 times, we can use the binomial probability formula:

`\[ P(X \leq k) = \sum_(i=0)^(k) \binom{n}{i} \cdot p^i \cdot (1-p)^(n-i) \]`

where:

n is the number of trials (rolls of the die),

k is the maximum number of successes we're interested in (at most 1 "one" in this case),

p is the probability of success on a single trial,

"n choose i" is the binomial coefficient, representing the number of ways to choose i successes out of n trials.

For a fair six-sided die, p (the probability of rolling a one on a single roll) is 1/6.

In this case, n is 5 (rolling the die 5 times) and k is 1 (at most 1 "one").

P(X <= 1) = (5 choose 0) * (1/6)^0 * (5/6)^5 + (5 choose 1) * (1/6)^1 * (5/6)^4

Let's calculate this:

P(X <= 1) = (5/6)^5 + 5 * (1/6) * (5/6)^4

P(X <= 1) is approximately 0.672.

So, rounded to the nearest thousandth, the probability of getting at most 1 "one" when rolling a fair six-sided die 5 times is approximately 0.672.

Complete question:

If a fair die is rolled 5 times, what is the probability, rounded to the nearest thousandth, of getting at most 1 one?