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Problem 10

AMC 10/12, early questions

Combinatorics Difficulty 3.0 Find the answer

There are $3$ students participating in a test, assuming that the probability of each student passing the test is $\frac{1}{3}$ , and whether each student passes the test is independent of each other. The probability that at least one student passes the test is ( )
A: $\frac{19}{27}$
B: $\frac{4}{9}$
C: $\frac{2}{3}$
D: $\frac{8}{27}$

Your answer

Official solution

To find the probability that at least one student passes the test, it's easier to first calculate the probability that none of them pass and then subtract that from $1$ .

1. The probability that a single student fails the test is $1 - \frac{1}{3} = \frac{2}{3}$ .
2. Since the students' performances are independent, the probability that all three students fail is $(\frac{2}{3})^3 = \frac{8}{27}$ .
3. Therefore, the probability that at least one student passes is $1 - \frac{8}{27} = \boxed{\frac{19}{27}}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.