probability?

Question

If there are 35 students in your class, and 4 students are chosen randomly to present projects each day until all projects have been presented, what is the probability that you will be chosen to make your presentation the first day?

If you have not been chosen by the 4th day, what is the probability that you will be chosen on the 4th day?

How do you set these up to solve them? I'm horrible at statistics....and need an explanation

Answer 1

Let me see if I can help the first day 4 students out of 35 are going so it would be 4/35
for each day 4 students go so by the 4th day that means 16 students have already gone(now we are on day 5)
so 35-16=19
SO there are 19 more students total left to present their project
so if 4 go each day the new probablity would be 4/19
HOWEVER and I am not too sure which way you want to do this so I will do both
by the 3rd day 12 people have gone(4 a day x 3 days)
then you tke the 35 -12 which gives you 23
so by the 4th day it would be 4/23
Does this help?

Answer 2

I am pretty sure the set-up looks like this---> 35! / 35!-4!
and then for the second part it will be 23! / 23!-4! If you've worked with these types of set-ups, then you shouldn't have a problem solving it. It's been a good 6 or 7 years since i have seen a problem like this, but i am pretty sure this is the set-up. Umm there are 35 students, and 4 are chosen, so you need to subtract the 4 from the 35 (therefore the 31!), and then since by the 4th day, there are only 23 students left (12 went on the first 3 days prior to the 4th), then you work with the 23 and subtract 4 again from that. I hope this helped a little...

Answer 3

well, the first part is 4 out of 35. that's the probability. it is set up as a ratio like this 4:35.

the best way to set it up is to create a diagram, with 4 hash marks in a box for each day, except the last day would have only 3 because 8 days would have 4 people.

then subtract 4 for each day that passes, becasue those students won't go again (thei is called "without replacement" - if they were to go again, you'd replace those students back into the original 35.

so, day two, you'd have a 4 out of 31 chance
day three, 4 out of 27 chance
day four, 4 out of 23 chance, and so on, subtracting the 4 for each day.