Respuesta :
Answer:
0.71 = 71% probability that she or he used the lab on a regular basis.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Student got an A
Event B: Used the lab on a regular basis.
Probability of an student getting an A:
55% of 40%(go to the lab on a regular basis).
15% of 100 - 40 = 60%(do not go to the lab on a regular basis).
So
[tex]P(A) = 0.55*0.4 + 0.15*0.6 = 0.31[/tex]
Probability of getting an A and using the lab on a regular basis:
55% of 40%, so:
[tex]P(A \cap B) = 0.55*0.4 = 0.22[/tex]
Probability that she or he used the lab on a regular basis.
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.22}{0.31} = 0.71[/tex]
0.71 = 71% probability that she or he used the lab on a regular basis.
