course bmitting Assignment

bms2216@email.vccs.edu

First off, I apologize for using the submit work form for my questions, but I could not locate a questions form. I wasn't sure if I should e-mail it or send it in a form.

This is the correct way to submit a question. However be sure to enter your access code in the box for access code, etc.. It was easy for me to correct that error, but if I hadn't noticed this wouldn't have posted correctly.

I have a couple of questions that I wasn’t sure how to do form one of the test I’ve printed off.

1) If 80% of scheduled flights actually take place and cancellation are independent event, what is the probability that 3 separate flights will take place?

The probability of each flight is 80% or .8, and the probabilities are independent. So the fundamental counting principle applies and the probability is .8 * .8 * .8 = .512.

2) Twenty percent of the students at Italie University are Itailan. If students are chosen at random, what is the probability that the 9th student will be Italian?

I wasn’t exactly sure which formula to use for these and I couldn’t find it in the book. I’m sure it’s there, but I couldn’t pick it out.

The probability is 20%. The probability of the 9th student being Italian is not affected by the unknown previous random choices.

A similar question would be a bit more complicated. If the question read

'Twenty percent of the students at Italie University are Itailan. If students are chosen at random, what is the probability that the 9th student will be first Italian chosen?'

then the first 8 students would all have to be non-Italian. For each student the probability would therefore be 80% = .8, and the probability for the first 8 students would be .8 ^ 8.

'Twenty percent of the students at Italie University are Itailan. If students are chosen at random, what is the probability that of the first 9 students, 3 will be Italian?'

This would be a binomial probability. There are C(9, 3) ways to distribute the 3 Italian students among the 9. For any such distribution, 3 will be Italian (probability .2 for each) and 6 will be non-Italian (probability .8 for each) so the probability will be .2^3 * .8^6.

The requested probability would then be C(9, 3) * .2^3 * .8^6.

In terms of successes and failures, which is the language of binomial probability, we would be looking at 3 successes in 9 tries, with probability of success being .2 and probability of failure being .8.