A survey of students revealed that 30% of them have a part-time job. If students are chosen at random what is the probability of the following events: That less than 2 have part time work? More than 3 are working part-time? None have a part-time job? This is a question from a mock exam paper as I prepare for exams. I guess I could start by imagining a number of 1000 college students surveyed with 300 having part-time jobs. so for part (i) -> (30/100)(70/10)(70/10)(70/10)(70/10)=7% That ain't right. Can someone shed some light on this?

Dulce Cantrell

Dulce Cantrell

Answered question

2022-09-14

A survey of students revealed that 30% of them have a part-time job. If students are chosen at random what is the probability of the following events: That less than 2 have part time work? More than 3 are working part-time? None have a part-time job?
This is a question from a mock exam paper as I prepare for exams.
I guess I could start by imagining a number of 1000 college students surveyed with 300 having part-time jobs.
so for part (i) -> (30/100)(70/10)(70/10)(70/10)(70/10)=7%
That ain't right. Can someone shed some light on this?

Answer & Explanation

Kiera Moreno

Kiera Moreno

Beginner2022-09-15Added 9 answers

As the 5 students are chosen at random, the probability for each of them of having a part-time job is still %. It furthermore has to assumed that there is independence among the students.If you let be the random variable that equals the number of students among the 5 chosen that have a part-time job, then as Jean-Claude Arbaut points out in his comment, has the binomial distribution .I'll show you how to find the probability of less than having a part-time job. Less than means at most , and we haveSince , we have and . Also, for any , and thusThis turns out to be approximately %.Edit: Allow me to make a comment about your approach: You suggest that the solution isThis is the same asorNotice how this is close to , which is . The reason why your answer is off by a factor of is that you consider only the case where the first student is the one with a part-time job, and where the remaining students don't have part-time jobs. It could also have been the second student, who had a part-time job, while student , , , and didn't. By changing which student has the part-time job, you getNotice how all five terms are just products of the form . Since there are of them, this gives you a total of .This shows you how you get from your idea to the correct expression for . But the question asks for the probability that there are less than students with a part-time job, so students with part-time jobs is also an option. This is why my approach above has an extra term compared to your suggestion.

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