Question

according to a poll, about 14% of adults in a country bet on professional sports. data indicate that 46.4% of the adult population in this country is male. complete parts (a) through (e).
a. yes. a person cannot both be male and bet on professional sports at the same time
b. yes. a person can both be male and bet on professional sports at the same time
c. no. a person can both be male and bet on professional sports at the same time.
d. no. a person cannot both be male and bet on professional sports at the same time
(b) assuming that betting is independent of sex, compute the probability that an adult from this country selected at random is a male and bets on professional sports.
p(male and bets on professional sports) = 0.0650
(type an integer or decimal rounded to four decimal places as needed.)
(c) using the result in part (b), compute the probability that an adult from this country selected at random is male or bets on professional sports.
p(male or bets on professional sports) = 0.5390
(type an integer or decimal rounded to four decimal places as needed.)
(d) the poll data indicated that 8.4% of adults in this country are males and bet on professional sports. what does this indicate about the assumption in part (b)? choose the correct answer.
a. the assumption was correct and the events are independent
b. the assumption was incorrect and the events are not independent

Explanation:

Step1: <Probability of independent events>

If two events \(A\) (being male) and \(B\) (betting on professional sports) are independent, the probability of both occurring is \(P(A\cap B)=P(A)\times P(B)\). Given \(P(A) = 0.464\) (probability of being male) and \(P(B)=0.14\) (probability of betting on professional sports), then \(P(A\cap B)=0.464\times0.14\).

$$0.464\times0.14 = 0.06496\approx0.0650$$

Step2: <Probability of union of two events>

The formula for \(P(A\cup B)\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Substitute \(P(A) = 0.464\), \(P(B)=0.14\), and \(P(A\cap B) = 0.0650\) into the formula.

$$P(A\cup B)=0.464 + 0.14-0.0650=0.5390$$

Step3: <Check independence assumption>

If the events were independent, \(P(A\cap B)\) (from part (b)) would be \(0.0650\) (or \(6.5\%\)). But the poll shows \(P(A\cap B) = 0.084\) (or \(8.4\%\)). Since \(0.0650
eq0.084\), the assumption of independence is wrong.

Answer:

For part (b): \(0.0650\)
For part (c): \(0.5390\)
For part (d): B. The assumption was incorrect and the events are not independent.