Bayes theorem and the inverse probabilities

Question

If I have the following data:

p(being in a fatal car accident) = 0.02%
p(driving while high) = 4.7%
p(driving while high | being in a fatal car accident) = 31.8%

I computed p(being in a fatal car accident | driving while high) = 0.135%

Is it possible to say:

p(driving while not high) = 1 - p(driving while high) = 95.3%?

and . . .

p(driving while not high | being in a fatal car accident) 
  = 1 - p(driving while high | being in a fatal car accident) 
  = 68.2%

Thus:

p(being in a fatal car accident | driving while not high) = 0.014%

Answer 1

DATA

$ \text{A} = \text{Being in a fatal car accident} = 0.0002 $

$ \text{B} = \text{Driving while high} = 0.047 $

$ P(B\mid A) = 0.318 $

GOAL

We want to calculate $P(A\mid \overline{B})$ = Being in a fatal car accident given that driving while not high.

PROCEDURE

As we want to calculate $P(A\mid \overline{B}) = \frac{P(A \cap \overline{B})}{P(\overline{B})}$, we need to obtain $P(A \cap \overline{B})$ and $P(\overline{B})$.

As we want to obtain $P(A \cap \overline{B})$, we proceed in this way:

$ P(B \mid A) = \frac{P(A \cap B)}{P(A)} $

$ P(A \cap B) = P(B \mid A) \cdot P(A) = 0.318 \cdot 0.0002 = 0.0000636 $

Since the probability of the intersection of an event with the complement of another event is the probability of the first event minus the probability of the intersection of the first event and the second:

$ P(A \cap \overline{B}) = P(A) - P(A \cap B) = 0.0002 - 0.0000636 = 0.0001364 $

As we want to obtain $P(\overline{B})$, we proceed in this way:

$ P(\overline{B}) = 1 - P(B) = 1 - 0.047 = 0.953 $

And now we calculate $P(A \mid \overline{B})$:

$ P(A \mid \overline{B}) = \frac{0.0001364}{0.953} = 0.000143 $

RESULT

The probability of being in a fatal car accident given that driving while not high is 0.0143%.