Question

1. the data on the right represent the number of traffic fatalities by seat location gender. determine the following: a. p(male) b. p(male|driver) c. are the events \male\ and \driver\ independent? show your work. d. are the events \male\ and \driver\ mutually exclusive? explain. e. p(passenger or female)

Explanation:

Step1: Calculate \( P(\text{male}) \)

The formula for probability is \( P(A)=\frac{n(A)}{n(S)} \). Here, \( n(\text{male}) = 39381 \) and \( n(S)=57600 \).
So, \( P(\text{male})=\frac{39381}{57600}= 0.6837 \)

Step2: Calculate \( P(\text{male}|\text{driver}) \)

By the formula of conditional probability \( P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{n(A\cap B)}{n(B)} \). Here, \( n(\text{male}\cap\text{driver}) = 6392 \) and \( n(\text{driver})=12753 \)
So, \( P(\text{male}|\text{driver})=\frac{6392}{12753}\approx0.501 \)

Step3: Check independence

Two events \( A \) (male) and \( B \) (driver) are independent if \( P(A\cap B)=P(A)\times P(B) \)
\( P(A)\times P(B)=0.6837\times\frac{12753}{57600}\)
\( P(A\cap B)=\frac{6392}{57600}\approx0.111 \)
\( 0.6837\times\frac{12753}{57600}\approx0.6837\times0.2214 = 0.1514
eq0.111 \)
So, the events “male” and “driver” are not independent.

Step4: Check mutual - exclusivity

Two events \( A \) and \( B \) are mutually exclusive if \( A\cap B=\varnothing \), i.e., \( n(A\cap B) = 0 \). Here, \( n(\text{male}\cap\text{driver})=6392
eq0 \)
So, the events “male” and “driver” are not mutually exclusive.

Step5: Calculate \( P(\text{passenger or female}) \)

By the formula \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \)
\( n(\text{passenger}) = 44847 \), \( n(\text{female})=18219 \), \( n(\text{passenger}\cap\text{female}) = 11858 \)
\( P(\text{passenger or female})=\frac{44847 + 18219-11858}{57600}=\frac{51208}{57600}\approx0.889 \)

Answer:

a. \( 0.6837 \)
b. \( 0.501 \)
c. No, since \( P(\text{male}\cap\text{driver})
eq P(\text{male})\times P(\text{driver}) \)
d. No, since \( n(\text{male}\cap\text{driver})
eq0 \)
e. \( 0.889 \)