53.41 is the answer to this
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
Answer:
3y + 6
Step-by-step explanation:
3 x y = 3y
3 x 2 = 6
3y + 6
Answer:
Step-by-step explanation:
There are 2 factors of 5, so you have 5^2
There are 3 factors of x, so you have x^3
There is a single y, so you have y
Answer:
Answer:
650
Step-by-step explanation: