Recent content by vvvidenov

so the event (student chosen is in grade 3) has size=7 so the probability is =1/7

ok, I see the response above.

Homework Statement A school contains students in grades 1,2,3,4,5 and 6. Grades 2,3,4,5 and 6 all contain the same number of students, but there are twice this number in grade 1. a)If a student is selected at random from a list of all the students in the school, what is the probability...

Thank you so much for your help. I get it now. I did not realize that I was doing wrong the entire problem, not just the words description.

Is it: BCc(blue cards numbered bigger than 5)={B6, B7...B10} the notation confuses me a lot: is it {BC6, BC7...BC10} or only {B6, B7...B10} I don't get it. Please explain it better, please.

I think is okay, b/c B={1,2,3...10} C={1,2,3,4} The BCcc={5,6...10}. It is the intersection B\capCc. and the Cc is all numbers except C.

Suppose that one card is to be selected from a deck of 20 cards taht cointains 10 red cards numbered from 1 to 10 and 10 blue cards numbered from 1 to 10. Let A be the event that a card with an even number is selected; let B be the event that the blue card is selected; and let C be the event...

Didn't you make a mistake under presure or in a hurry to get many thinks done. I mention that I am almost 50 years old and I am back to school. I it is diffucult for me. I can get this problem done even without your help. I don't blame anybody. But similar example was given and you have no...

Mark, I agree with you. I read in the book a lot.

Thank you for the proof of my work. That is the way the professor wants us to do it. :smile:

How large is k+1? n=k=4 k+1=5 Then what?

I am not sure , but I will try. 6^k > 2^k*2 I don't think I can finish this problem. I asked in class similar problem ,but this professor is happy to make us more confused, he doesn't explain well at all. I also don't think that other kids in the class get it. I just don't get it, sorry.

I am sorry, thanks for the note. I know that I have to start first with the proof in base part, I just skip it in a hurry. :redface: Here is: let n=4 n!>2^n 4!>2^4 24>16 (base) n=k k!>2^(k+1) consider for n=k+1 (k+1)!= (k+1)k!>(k+1)(2^k) I don't know how to show it. Please...

for part b) from above: 2m^6< n^4 2m^6 < n^4 multiply both sides by 81 81(2m^6) < 81n^4 162m^6 < (3n)^4 128m^6 < 162m^6 2*64=128m^6<81n^4 2(2m)^6 < (3n)^4 2^6=64 and 3^4=81 this is how they find 64 and 81 and start with them but nothing is clear. My problem is combined...

Mark, thank you very much for showing me this way. I will work on this problem again and I will be glad if you review it for me again. It is complicated problem. I really want to find how to solve it. I am not surprise that no one else try it. :smile: 81 and 64 I have from solving (3m)^5 and...