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) are the events \male\ and \bet on professional sports\ mutually exclusive? explain. choose the correct answer
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) =
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Recall the formula for \(P(A\cup B)\)

The formula for the probability of the union of two events \(A\) and \(B\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Let \(A\) be the event "male" and \(B\) be the event "bets on professional sports". We know that \(P(A) = 0.464\), \(P(B)=0.14\), and from part (b) \(P(A\cap B)=0.0650\).

Step2: Substitute the values into the formula

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

Step3: Calculate the result

First, add \(0.464\) and \(0.14\): \(0.464+0.14 = 0.604\). Then subtract \(0.0650\) from \(0.604\): \(0.604-0.0650=0.539\).

Answer:

\(0.5390\)