PLSSS HLEPP ME OUTT ITS URGENTTTT!!
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Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.
Conditional Probability
In which
- P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
- Event A: Fail the test.
- Event B: Unfit.
The probability of <u>failing the test</u> is composed by:
- 46% of 37%(are fit).
- 100% of 63%(not fit).
Hence:
The probability of both failing the test and being unfit is:
Hence, the conditional probability is:
0.7873 = 78.73% probability that Mona was justifiably dropped.
A similar problem is given at brainly.com/question/14398287