View Full Version : Stats Question, need HELP please!!!


iJackTeeth
12-04-2008, 10:52 AM
John was interviewed at 4 schools, the interview to acceptance percentage at the four schools were 45%, 35%, 70%, and 56%, what is johns chance of getting at least one acceptance to a school?

Streetwolf
12-04-2008, 12:42 PM
John was interviewed at 4 schools, the interview to acceptance percentage at the four schools were 45%, 35%, 70%, and 56%, what is johns chance of getting at least one acceptance to a school?
Say the chance was 80% (it probably isn't). That means the chance he was accepted to NONE of the schools is 20%. So we want to solve for the chances he wasn't accepted anywhere and then figure out the chances he got at least 1 acceptance.

P(no acceptance) = 55% * 65% * 30% * 44% (these are the percentages obtained when you do (100 - x) to all four percentages above - they represent the percentage of not getting accepted after an interview)

P(no acceptance) = (0.55)(0.65)(0.30)(0.44) = 0.04719
P(no acceptance) = 4.719%
P(acceptance somewhere) = 100 - 4.719 = 95.381%

Sublimation
12-04-2008, 12:45 PM
Wats the answer? is it 51.5 percent ????

Sublimation
12-04-2008, 12:48 PM
Say the chance was 80% (it probably isn't). That means the chance he was accepted to NONE of the schools is 20%. So we want to solve for the chances he wasn't accepted anywhere and then figure out the chances he got at least 1 acceptance.

P(no acceptance) = 55% * 65% * 30% * 44% (these are the percentages obtained when you do (100 - x) to all four percentages above - they represent the percentage of not getting accepted after an interview)

P(no acceptance) = (0.55)(0.65)(0.30)(0.44) = 0.04719
P(no acceptance) = 4.719%
P(acceptance somewhere) = 100 - 4.719 = 95.381%

oooohhhh,:idea: now that makes more sense. thanks.

IWantSmile
12-07-2008, 09:05 PM
Say the chance was 80% (it probably isn't). That means the chance he was accepted to NONE of the schools is 20%. So we want to solve for the chances he wasn't accepted anywhere and then figure out the chances he got at least 1 acceptance.

P(no acceptance) = 55% * 65% * 30% * 44% (these are the percentages obtained when you do (100 - x) to all four percentages above - they represent the percentage of not getting accepted after an interview)

P(no acceptance) = (0.55)(0.65)(0.30)(0.44) = 0.04719
P(no acceptance) = 4.719%
P(acceptance somewhere) = 100 - 4.719 = 95.381%

It makes so much sense when you explain in this. :rolleyes: