Probability of Randomly Selective Event, Conditional Probability

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The discussion revolves around calculating probabilities related to a company's advertisement and product purchase behavior. The probability of an individual buying the product is established at 20% after considering the advertisement's influence. For part B, to find the probability that at least one out of five randomly selected individuals will buy the product, the approach involves calculating the probability that none buy, which is (1 - 0.2)^5. This leads to the conclusion that the probability of at least one purchase is 1 minus the probability that none buy. The calculations emphasize the use of conditional probability and the binomial distribution in determining outcomes.
conniebear14
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Homework Statement



A company has been running a television advertisement for one of its new products. A survey was conducted. Based on its results, it was concluded that an individual buys the product with probability https://utdvpn.utdallas.edu/wwtmp/equations/42/,DanaInfo=.aevnhvE00ljvwm5Ntt.,SSL+b76747aa0afb1816e5979c66ce77851.png , if he/she saw the advertisement, and buys with probability https://utdvpn.utdallas.edu/wwtmp/equations/1e/,DanaInfo=.aevnhvE00ljvwm5Ntt.,SSL+e895ee9ca85bcbf1e55a96a7573c291.png , if he/she did not see it. Twenty-five percent of people saw the advertisement.

a. What is the probability that a randomly selected individual will buy the new product?
b. What is the probability that at least one of randomly selected five individuals will buy the new product?

Homework Equations


P(A|B) = P(B|A)P(A)/P(B)

The Attempt at a Solution


I already got part A correct.
The answer is .2
I am confused on part B probably because of the 1/5 thing. Which equation should I use and where should I start with this one?[/B]
 
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it didn't post the numbers but the blanks correspond to 56% and 8% respectively
 
You have the probability a given person buys is ##.2##. What is the probability they all fail to buy? You could use the binomial distribution but it is easy enough to just calculate.
 
LCKurtz said:
You have the probability a given person buys is ##.2##. What is the probability they all fail to buy? You could use the binomial distribution but it is easy enough to just calculate.
Okay so I calculated probability person does not buy as .8 from (1-.56)(.25) + (.92)(.75). But now I am stuck. Where does the five part come in? What should I do next?
 
If the probability the first person doesn't buy is ##.8##, and if they are independent, what is the probability the next person doesn't buy? So...
 
conniebear14 said:

Homework Statement



A company has been running a television advertisement for one of its new products. A survey was conducted. Based on its results, it was concluded that an individual buys the product with probability 58%, if he/she saw the advertisement, and buys with probability 8% if he/she did not see it. Twenty-five percent of people saw the advertisement.

a. What is the probability that a randomly selected individual will buy the new product?
b. What is the probability that at least one of randomly selected five individuals will buy the new product?

Homework Equations


P(A|B) = P(B|A)P(A)/P(B)

The Attempt at a Solution


I already got part A correct.
The answer is .2
I am confused on part B probably because of the 1/5 thing. Which equation should I use and where should I start with this one?[/B]
If the answer to (a), which you say you got, is p, then the probability that "at least one" will buy is the 1 minus the probability none will buy. The probability that none will buy is (1- p)^5
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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