Question

a study showed that 61.5% of occupants involved in a fatal car crash wore seat belts. of those in a fatal car crash who wore seat belts, 3% were ejected from the vehicle. for 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
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Define given probabilities

Let \( S \) = event of wearing a seatbelt, \( E \) = event of being ejected.
\( P(S) = 0.615 \), \( P(
eg S) = 1 - 0.615 = 0.385 \)
\( P(E|S) = 0.03 \), \( P(E|
eg S) = 0.36 \)

Step2: Calculate total ejection probability

Use law of total probability:
\( P(E) = P(E|S)P(S) + P(E|
eg S)P(
eg S) \)
\( P(E) = (0.03 \times 0.615) + (0.36 \times 0.385) \)
\( P(E) = 0.01845 + 0.1386 = 0.15705 \)

Step3: Apply Bayes' Theorem

Find \( P(S|E) = \frac{P(E|S)P(S)}{P(E)} \)
\( P(S|E) = \frac{0.03 \times 0.615}{0.15705} \)

Step4: Compute final value

\( P(S|E) = \frac{0.01845}{0.15705} \approx 0.1175 \)

Answer:

0.1175