Working on summer vacation. Is summer vacation a break from work? Not according to a Harris Interactive (July 2013) poll of U.S. adults. The poll found that 61% of the respondents work during their summer vacation, 22% do not work while on vacation, and 17% are unemployed. Assuming these percentages apply to the population of U.S. adults, consider the work status during the summer vacation of a randomly selected adult.

a. What is the Probability that the adult works while on summer vacation?

b. What is the Probability that the adult will not work while on summer vacation, either by choice or due to unemployment?

Short Answer

Expert verified
  1. P(W)=0.61
  2. P(donotworkorunemployed)=0.39

Step by step solution

01

Step-by-Step SolutionStep 1: Identify the Probability of the adult work while on summer vacation

The number of events in an exhaustive collection of equally likely outcomes that result in a particular occurrence divided by the total number of conceivable outcomes is known as Probability.

Let’s assume:

W =Respondents work during the summer vacation.

N = Respondents do not work during the summer vacation.

U = Respondents were unemployed.

Therefore, from the given data, we get:

W=61%N=22%U=17%

The probabilities of work during the summer vacation:

P(W)=61%=61100=0.61

Hence, the Probability of who was working during summer vacation is0.61.

02

Identify the Probability that the respondent will not work during summer vacation, either by choice or by unemployed

The Probability of respondents not working during the summer vacation:

P(N)=22%=22100=0.22

The Probability of respondent being unemployed:

P(U)=17%=17100=0.17

Therefore, the Probability of respondents does not work or being unemployed:

role="math" localid="1653479154719" P(donotworkorunemployed)=P(N)+P(U)=0.22+0.17=0.39

Hence, the probability of not working or being unemployed during summer vacation is0.39.

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Most popular questions from this chapter

Suppose the events B1,B2,B3 are mutually exclusive and complementary events, such thatP(B1)=0.2, P(B2)=0.4and P(B3)=0.5. Consider another event A such thatP(AB1)=P(AB2)=0.1andP(AB3)=0.2Use Baye’s Rule to find

a.P(B1A)

b.PB2A

c.role="math" localid="1658214716845" P(B3A)

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