Respuesta :
Answer:
A. [tex]P(\frac{Br}{B})= \frac{0.22}{0.36}[/tex]
Step-by-step explanation:
Conditional probability:-
If E and F are two events in a sample space S and P( F) ≠ 0 , then the probability of 'F' after the event 'E' has occurred, is called the conditional probability of the event of F given E and is denoted by
P(E/F) = [tex]\frac{P(E n F)}{P(F)}[/tex]
"Police report that 78% of drivers are given a breath test
36% a blood test and 22% both tests
let 'B' be the event of blood test and 'Br' be the event of breath test
[tex]P(\frac{Br}{B})= \frac{P(Br n B)}{P(B)}[/tex]
[tex]P(\frac{Br}{B})= \frac{22/100}{36/100)}[/tex]
[tex]P(\frac{Br}{B})= \frac{0.22}{0.36}[/tex]
Answer:
Probability of (breath test / blood test) = 0.22/0.36.
Step-by-step explanation:
We are given that Police report that 78% of drivers are given a breath test, 36% a blood test, and 22% both tests.
Let Probability that drivers are given a breath test = P(X) = 0.78
Probability that drivers are given a blood test = P(Y) = 0.36
Probability that drivers are given both tests = [tex]P(X \bigcap Y)[/tex] = 0.22
Now, we have to find Probability of (breath test/blood test);
As we know that Conditional Probability of (A/B) is given by;
P(A/B) = [tex]\frac{P(A \bigcap B)}{P(B)}[/tex]
Similarly, P(breath test/blood test) = P(X/Y) = [tex]\frac{P(X \bigcap Y)}{P(Y)}[/tex]
= [tex]\frac{0.22}{0.36}[/tex] = 0.61
Therefore, required probability is 0.22/0.36 = 0.61.
