Prob(Math OR English)
= P(M) + P(E) - P(M and E)
= .8 + .9 - .72
= .98
Prob (failing both) = 1 - .98
= .02
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
2 answers
Thanks