Answer:
1/16
Step-by-step explanation:
(4/16)^2 = 1/16. Input the following in the calculator.
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:
Both (b) and (C) are true.
This study is double blind and the "only spice tea" group serves as the control group.
A. 56 cards
B. 120 cards
Sorry if it's wrong, I'm kind of confused. :P
Answer: 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
4: 4, 8, 12, 16, 20, 24, 28