A teacher administers three different versions of a test to his students. The two-way table shows the results. Find each probability.Grade Pass FailTestA 49 7B 46 6C 34 12a. P(pass | Test A)Probability: $\%$%b. P(Test C | fail)Probability: $\%$
Question
A teacher administers three different versions of a test to his students. The two-way table shows the results. Find each probability.Grade Pass FailTestA 49 7B 46 6C 34 12a. P(pass | Test A)Probability: %b. P(Test C | fail)Probability:
Solution
a. P(pass | Test A)
This is the probability of passing given that the student took Test A.
To find this, we divide the number of students who passed Test A by the total number of students who took Test A.
Number of students who passed Test A = 49 Total number of students who took Test A = 49 (passed) + 7 (failed) = 56
So, P(pass | Test A) = 49 / 56 = 0.875
Therefore, the probability is 87.5%
b. P(Test C | fail)
This is the probability that a student took Test C given that they failed.
To find this, we divide the number of students who failed Test C by the total number of students who failed.
Number of students who failed Test C = 12 Total number of students who failed = 7 (Test A) + 6 (Test B) + 12 (Test C) = 25
So, P(Test C | fail) = 12 / 25 = 0.48
Therefore, the probability is 48%
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