At a high school, 30% of the students buy class rings. You
select five students at random. Find P(at least four buy
rings).
**Please include 5 decimal places in your answer**
This is a binomial probability problem, where each student is a trial with a success (buying a class ring) probability of 0.3. We want to find the probability of having at least four successes in five trials, which is equal to:
P(X >= 4) = P(X = 4) + P(X = 5)
Using the binomial probability formula, we get:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where n is the number of trials, k is the number of successes, and p is the probability of success.
P(X = 4) = (5 choose 4) * 0.3^4 * 0.7^1 = 0.072027
P(X = 5) = (5 choose 5) * 0.3^5 * 0.7^0 = 0.00243
Therefore,
P(X >= 4) = 0.072027 + 0.00243 = 0.074457
Rounded to 5 decimal places, the probability that at least four out of five selected students will buy class rings is 0.07446.
Need to the find the probability of at least 4 out of 5 and 5 out of 5 since the question asks you to find at least 4 buy rings out of 5 students.
nCr(5,4)(0.30)^4(0.70)^1 + nCr(5,5)(0.30)^5(0.70)^0 = 0.03078
Your calculation is incorrect. The probability of at least 4 out of 5 and 5 out of 5 are not added together, but rather the probabilities are calculated separately and then added.
The probability of exactly 4 out of 5 students buying rings is:
P(X = 4) = nCr(5,4) * 0.3^4 * 0.7^1 = 0.07203
The probability of exactly 5 out of 5 students buying rings is:
P(X = 5) = nCr(5,5) * 0.3^5 * 0.7^0 = 0.00243
To find the probability of at least 4 out of 5 students buying rings, we add the probabilities of having 4 and 5 students buying rings:
P(X >= 4) = P(X = 4) + P(X = 5) = 0.07203 + 0.00243 = 0.07446
Therefore, the probability of at least 4 out of 5 students buying class rings is 0.07446. This is the final answer, rounded to 5 decimal places.
To find the probability of at least four students buying rings, we need to consider two scenarios: exactly four students buying rings and exactly five students buying rings.
First, we need to find the probability of exactly four students buying rings. Since the probability of a single student buying a ring is 30%, the probability of a single student not buying a ring is 1 - 0.30 = 0.70. Therefore, the probability of exactly four students buying rings can be calculated using the binomial probability formula:
P(exactly four students buying rings) = C(n, k) * p^k * (1 - p)^(n - k)
where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and C(n, k) is the number of combinations.
In this case, we have n = 5 (five students), k = 4 (four students buying rings), and p = 0.30 (probability of a single student buying a ring).
P(exactly four students buying rings) = C(5, 4) * (0.30)^4 * (0.70)^(5 - 4)
Now, let's calculate it step by step.
C(5, 4) = 5! / (4! * (5 - 4)!) = 5
P(exactly four students buying rings) = 5 * (0.30)^4 * (0.70)^1
= 5 * 0.081 * 0.70
= 0.2835
Next, we need to find the probability of exactly five students buying rings. Using the same formula, we have n = 5, k = 5, and p = 0.30.
P(exactly five students buying rings) = C(5, 5) * (0.30)^5 * (0.70)^(5 - 5)
Now, let's calculate it step by step.
C(5, 5) = 5! / (5! * (5 - 5)!) = 1
P(exactly five students buying rings) = 1 * (0.30)^5 * (0.70)^0
= 1 * 0.00243 * 1
= 0.00243
Finally, to find the probability of at least four students buying rings, we sum the probabilities of exactly four and exactly five students buying rings:
P(at least four students buying rings) = P(exactly four students buying rings) + P(exactly five students buying rings)
= 0.2835 + 0.00243
= 0.28593 (rounded to 5 decimal places)
Therefore, the probability of at least four students buying rings is approximately 0.28593.