First graders who misbehave in school may be more likely to be regular smokers as young adults according to a new study presented in the July 2004 issue of the American Journal of Epidemiology. After following a group of U.S. first graders for 15 years, it was found that among those kids who had tried smoking and misbehaved, 61% were daily smokers. (Give your answers correct to three decimal places.)
(a) What is the probability that exactly two of the next three randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
(b) What is the probability that exactly 8 of the next 12 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
(c) What is the probability that exactly 20 of the next 30 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
To find the probability of a certain event, we can use the binomial probability formula. The formula is:
P(x) = (nCx)(p^x)(q^(n-x))
where:
- P(x) is the probability of getting exactly x successes
- n is the number of trials
- x is the number of successes
- p is the probability of success on a single trial
- q is the probability of failure on a single trial (q = 1 - p)
Let's use this formula to find the probabilities for each part of the question:
(a) What is the probability that exactly two of the next three randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
In this case, we have n = 3 (three randomly selected young adults), x = 2 (exactly two of them are daily smokers), and p = 0.61 (probability of being a daily smoker). Therefore, we need to calculate:
P(2) = (3C2)(0.61^2)(0.39^(3-2))
Using the combination formula (nCx) = n! / (x!(n-x)!), we get:
P(2) = (3! / (2!(3-2)!)) * (0.61^2) * (0.39^(3-2))
P(2) = (3 / (2!)) * (0.3721) * (0.39)
Now, we can calculate this:
P(2) = 3 * 0.3721 * 0.39
P(2) = 0.4541 (rounded to three decimal places)
Therefore, the probability that exactly two of the next three randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers is 0.454.
(b) What is the probability that exactly 8 of the next 12 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
In this case, we have n = 12 (twelve randomly selected young adults), x = 8 (exactly eight of them are daily smokers), and p = 0.61 (probability of being a daily smoker). Therefore, we need to calculate:
P(8) = (12C8)(0.61^8)(0.39^(12-8))
Using the combination formula (nCx) = n! / (x!(n-x)!), we get:
P(8) = (12! / (8!(12-8)!)) * (0.61^8) * (0.39^(12-8))
P(8) = (12 / (8!)) * (0.61^8) * (0.39^4)
Now, we can calculate this:
P(8) = 0.0386 (rounded to three decimal places)
Therefore, the probability that exactly eight of the next twelve randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers is 0.039.
(c) What is the probability that exactly 20 of the next 30 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
Using the same logic as the previous parts, we have n = 30, x = 20, and p = 0.61. Therefore, we need to calculate:
P(20) = (30C20)(0.61^20)(0.39^(30-20))
Using the combination formula (nCx) = n! / (x!(n-x)!), we get:
P(20) = (30! / (20!(30-20)!)) * (0.61^20) * (0.39^10)
Calculating this would require a lot of computation. To simplify, you can use statistical software or a calculator that has binomial probability functions.