Betting on Sports According to a Gallup Poll, about 17% of adult Americans bet on professional sports. Census data indicate that 48.4% of the adult population in the United States is male.

(a) Assuming that betting is independent of gender, compute the probability that an American adult selected at random is a male and bets on professional sports.
(b) Using the result in part (a), compute the probability that an American adult selected at random is male or bets on professional sports.
(c) The Gallup poll data indicated that 10.6% of adults in the United States are males and bet on professional sports. What does this indicate about the assumption in part (a)?
(d) How will the information in part (c) affect the probability you computed in part (b)?

Solution

Let $A$ denote the event that an adult Americans bet on professional sports. Let $B$ denote the event that the adult population in the United States is male.

About 17% of adult Americans bet on professional sports, i.e. $P(A) =0.17$.

Census data indicate that 48.4% of the adult population in the United States is male, i.e., $P(B)=0.484$.

a. Assuming that betting is independent of gender, the probability that an American adult selected at random is a male and bets on professional sports is

$$ \begin{aligned} P(A \text{ and } B) &= P(A\cap B)\\ &=P(A)P(B)\\ &= 0.17*0.484\\ &=0.08228 \end{aligned} $$

b. The probability that an American adult selected at random is male or bets on professional sports is

$$ \begin{aligned} P(A \text{ OR } B) &= P(A\cup B)\\ &=P(A)+P(B)-P(A\cap B)\\ &= 0.17+0.484- 0.08228\\ &=0.57172 \end{aligned} $$

c. The Gallup poll data indicated that 10.6% of adults in the United States are males and bet on professional sports.

Given that $P(A\cap B) = 0.106$

But according to part (a) $P(A)*P(B) = 0.17\times *0.484 = 0.08228$.

Thus $P(A\cap B) \neq P(A)P(B)$, $A$ and $B$ are not independent.

d. The probability that an American adult selected at random is male or bets on professional sports using informaton in (c) is

$$ \begin{aligned} P(A \text{ OR } B) &= P(A\cup B)\\ &=P(A)+P(B)-P(A\cap B)\\ &= 0.17+0.484- 0.106\\ &=0.548 \end{aligned} $$

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