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I have a probability question (please read below this to see actual question). Can someone help and explain?

A survey found that 30% of teenagers received spending money from part-time jobs. If five teens are selected at random, find the probability that at least 3 will have part-time jobs. So - the first question you need to ask, what is n - the number of trials? Then - what is the probability of success (finding a teenage who has a p-t job)? Finally - what is x, the number of successes that we are looking for? This gets a little more complicated because it is not a single value here. Look at the wording - "at least 3". What are the possible values of x?

Public Comments

1. isnt that the type of problem that you have to use a random number chart..... so then 0-30 would be part time jobs and 31-100 would not have part time jobs. so you circle numbers that are 0-30 and see how many in each row and then see if there are at least 3 (3 or more) hope that helped, if not, heres a link that shows how to do that type of problem... http://my.hrw.com/tabnav/controller.jsp?isbn=0030784867 im not sure wich videos but i know theyre there and its the ones with the random number charts. hope this helped
2. binomial distribution n = 5 P(X =a teenage has a p-t job) = probability of success is .3 P(X ≤ 3) = P(X > 3) = 1 - [P(X = 4) + P(X = 5)] = 1 - [5C4 (.3)^4*(.7) + 5C5(.3)^5] = 1 - [.02835 + .00243] = .96922
3. The number of trials is 5. Probability of finding one teenager with a p-t job is 0.3. x=3, 4, & 5. P(x≥3)= P(x=3)+P(x=4)+P(x=5) P(x=3)=5C3*0.3^3=10*0.027=0.27 P(x=4)=5C4*0.3^4=5*0.0081=0.0405 P(x=5)=5C5*0.3^5=1* 0.00243 P(x≥3)=0.31293 ANSWER