Question

a study showed that 61.5% of occupants involved in a fatal car crash wore seat belts. of those not wearing seat belts, 36% were ejected from the vehicle. complete parts (a) and (b) below.
(a) find the probability that a randomly selected person in a fatal car crash who was ejected from the vehicle was wearing a seatbelt. the probability is 0.1175. (type an integer or decimal rounded to four decimal places as needed.)
(b) find the probability that a randomly selected person in a fatal car crash who was not ejected from the vehicle was not wearing a seatbelt. the probability is \\(\square\\). (type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Define given probabilities

Let $P(W) = 0.615$ (probability of wearing seat belt), $P(
eg W) = 1 - 0.615 = 0.385$ (probability of not wearing seat belt).
$P(E|W) = 0.03$ (ejected given seat belt worn), $P(E|
eg W) = 0.36$ (ejected given no seat belt).

Step2: Calculate total ejected probability

Use law of total probability:

$$P(E) = P(E|W)P(W) + P(E| eg W)P( eg W)$$

$$P(E) = (0.03 \times 0.615) + (0.36 \times 0.385) = 0.01845 + 0.1386 = 0.15705$$

Step3: Solve part (a) with Bayes' theorem

Find $P(W|E)$:
$$P(W|E) = \frac{P(E|W)P(W)}{P(E)}$$
$$P(W|E) = \frac{0.03 \times 0.615}{0.15705} = \frac{0.01845}{0.15705} \approx 0.1175$$

Step4: Solve part (b) with complement rule

First find $P(
eg E|
eg W) = 1 - P(E|
eg W) = 1 - 0.36 = 0.64$
Then use total probability for non-ejected:

$$P( eg E) = 1 - P(E) = 1 - 0.15705 = 0.84295$$

Use Bayes' theorem for $P(
eg W|
eg E)$:

$$P( eg W| eg E) = \frac{P( eg E| eg W)P( eg W)}{P( eg E)}$$

$$P( eg W| eg E) = \frac{0.64 \times 0.385}{0.84295} = \frac{0.2464}{0.84295} \approx 0.2923$$

Answer:

(a) 0.1175
(b) 0.2923