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
Step-by-step explanation:
1 2/3 can be written as 5/3 and -3 1/2 can be written as -7/2
now,
→ 5/3×-7/2
→ 5 × (-7) / 3 × 2
→ -35/6 ans.
hope this answer helps you dear...take care and may u have a great day ahead!
Answer:
i think its b.
Step-by-step explanation:
correct me if im wrong. and it was worth 10 points instead of 20.
Answer:
$30.69
Step-by-step explanation:
22.32/ 8 = 2.79
11 x 2.79 = 30.69