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amid [387]
3 years ago
10

2²ˣ=8 I have a list of work to do and i need help

Mathematics
1 answer:
gayaneshka [121]3 years ago
4 0

Answer:

3/2

Step-by-step explanation:

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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

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) 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:

P(A) = 0.46(0.37) + 0.63 = 0.8002

The probability of both failing the test and being unfit is:

P(A \cap B) = 0.63

Hence, the conditional probability is:

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.63}{0.8002} = 0.7873

0.7873 = 78.73% probability that Mona was justifiably dropped.

A similar problem is given at brainly.com/question/14398287

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2 years ago
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