Mary's class took two tests last week. 80% of the class passed the math test, 90% of the class passed the English test, and 72% of the class passed both. What is the probablility that a randomly selected student in Mary's class failed both tests? Express your answer as a percent rounded to the nearest tenth of a percent?

Question

Mary's class took two tests last week. 80% of the class passed the math test, 90% of the class passed the English test, and 72% of the class passed both. What is the probablility that a randomly selected student in Mary's class failed both tests? Express your answer as a percent rounded to the nearest tenth of a percent?

Thanks
-Zach

Answer 1

Prob(Math OR English)

= P(M) + P(E) - P(M and E)
= .8 + .9 - .72
= .98

Prob (failing both) = 1 - .98
= .02

Answer 2

To find the probability that a randomly selected student in Mary's class failed both tests, we need to determine the percentage of students who did not pass either the math or English test.

First, let's calculate the percentage of students who passed at least one of the tests. We can do this by subtracting the percentage of students who failed both tests from 100%:

Percentage passed at least one test = 100% - 72% = 28%

Since we want the probability of failing both tests, we subtract the percentage passed at least one test from 100%:

Probability of failing both tests = 100% - 28% = 72%

So, the probability that a randomly selected student in Mary's class failed both tests is 72%. Rounded to the nearest tenth of a percent, it is 72.0%.