What is the probability that a randomly selected worker signed up for both class

Question

What is the probability that a randomly selected worker signed up for both classes? What is the probability that a randomly selected worker who signed up for the mathematics classes also signed up for the reading classes? What is the probability that a randomly chosen worker signed up for at least one of these two classes? Are the events "signs up for the reading classes" and "signs up for the mathematics classes" statistically independent? A lawn-care service makes telephone solicitations, seeking customers for the coming season. A review of the records indicates that 15% of these solicitations produce new customers and that, of these new customers, 80% had used some rival service in the previous year. It is also estimated that, of all solicitation calls made, 60% are to people who had used a rival service the previous year. What is the probability that a call to a person who had used a rival service the previous year will produce a new customer for the lawn-care service? An editor may use all, some, or none of three possible strategies to enhance the sales of a book: An expensive prepublication promotion An expensive cover design A bonus for sales representatives who meet predetermined sales levels In the past, these three strategies have been applied simultaneously to only 2% of the company's books Twenty percent of the books have had expensive cover designs, and, of these, 80% have had expensive prepublication promotion. A rival editor learns that a new book is to have both an expensive prepublication promotion and an expensive cover design and now wants to know how likely it is that a bonus scheme for sales representatives will be introduced. Compute the probability of interest to the rival editor.

Explanation / Answer

Here we have

P(new customers)=0.15

P(previous year rival service|new customers) = 0.80

P(previous year rival service) = 0.60

We need to find P(new customer |previous year rival service) =??

From the conditional probability we have

P(previous year rival service and new customers)=P(previous year rival service|new customers) P(new customers)= 0.80 *0.15 = 0.12

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And again from the conditional probability we have

P(new customer |previous year rival service) = P(previous year rival service and new customers) / P(previous year rival service) = 0.12/ 0.60 = 0.2

Hence, required probability is 0.2.

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