Question 1198291: Approximately 10.3% of
American high school students drop out of school before
graduation. Choose 10 students entering high school at
random. Find the probability that
a. No more than two drop out
b. At least 6 graduate
c. All 10 stay in school and graduate
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **a) Probability that no more than two drop out**
* This means we need to find the probability that 0, 1, or 2 students drop out.
* We can use the binomial probability formula:
P(X = k) = (nCk) * (p^k) * ((1-p)^(n-k))
where:
* n = number of trials (10 students)
* k = number of successes (dropouts)
* p = probability of success (dropout rate = 0.103)
* (nCk) = n! / (k! * (n-k)!)
* P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
* Calculate each probability and sum them:
* P(X = 0) = (10C0) * (0.103)^0 * (0.897)^10 ≈ 0.3372
* P(X = 1) = (10C1) * (0.103)^1 * (0.897)^9 ≈ 0.3708
* P(X = 2) = (10C2) * (0.103)^2 * (0.897)^8 ≈ 0.2015
* P(X ≤ 2) ≈ 0.3372 + 0.3708 + 0.2015 ≈ 0.9095
**Therefore, the probability that no more than two students drop out is approximately 0.9095.**
**b) Probability that at least 6 graduate**
* This means the probability that 4 or fewer students drop out (since 10 - 4 = 6).
* P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
* Calculate each probability and sum them:
* P(X = 3) = (10C3) * (0.103)^3 * (0.897)^7 ≈ 0.0574
* P(X = 4) = (10C4) * (0.103)^4 * (0.897)^6 ≈ 0.0118
* P(X ≤ 4) ≈ 0.9095 + 0.0574 + 0.0118 ≈ 0.9787
* P(at least 6 graduate) = 1 - P(X ≤ 4) = 1 - 0.9787 ≈ 0.0213
**Therefore, the probability that at least 6 students graduate is approximately 0.0213.**
**c) Probability that all 10 stay in school and graduate**
* This means that 0 students drop out.
* P(X = 0) = (10C0) * (0.103)^0 * (0.897)^10 ≈ 0.3372
**Therefore, the probability that all 10 students stay in school and graduate is approximately 0.3372.**
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