What needs to be calculated is her "true" correct answer rate. That is, assuming that the odds of her passing with 771 tries are 50%, and assuming that she just chooses a random answer each time she doesn't know, what percentage of answers does she know?

edit: On second thought, couldn't she just have used the $3,600 to bribe someone for a passing grade?

NathanSC

United States620 Posts

February 06 2009 09:52 GMT

#82

The odds of passing:

Assuming 10 questions with 4 multiple choice answers.

In order to pass the test, the woman must answer at least 6/10 questions correctly. This presents 5 distinct possibilities for passing the test. She may pass with 60%, 70%, 80%, 90%, or 100%. Furthermore, the woman has a .25 chance of getting an answer correct and a .75 chance of getting an answer incorrect.

The likelihood of getting each individual percentage is determined by the following:

( .25^(# correct questions desired) * .75^(# of incorrect questions desired) ) * 10! / ( (#number of correct questions desired) ! * (#number of incorrect questions desired) ! )

More simply, for the first option of 100%, you have:

( .25^10 * .75^0 ) * 10! / (10! * 0!)
( .25^10 * 1 ) * 10! / (10! * 1)
( .25^10 ) * 10! / 10!
( .25^10 )
.25 * .25 * .25 * .25 * .25 * .25 * .25 * .25 * .25 * .25

The likelihood of her getting each individual percentage needs to be determined then added together. A quick explanation of the math is that the percent chance of receiving a % grade is determined by your chances of getting each answer correct or incorrect then multiplied by the number of possible iterations.

As anyone can see, for 100% the possible arrangements of the .25 are limited to only 1. With 90% you have a still obvious 10 arrangements of the .75. With 80% there are 45, and so forth.

(.25 * .25 * .25 * .25 * .25 * .25 * .25 * .25 * .25 * .25) * 1
(.25 * .25 * .25 * .25 * .25 * .25 * .25 * .25 * .25 * 75) * 10
(.25 * .25 * .25 * .25 * .25 * .25 * .25 * .25 * .75 * .75) * 45
(.25 * .25 * .25 * .25 * .25 * .25 * .25 * .75 * .75 * .75) * 120
(.25 * .25 * .25 * .25 * .25 * .25 * .75 * .75 * .75 * .75) * 210

Multiplying gives you:

9.53674*10^-7 + .000029 + .000386 + .00309 + .016222 = .019728

This is a 1.97277% chance of passing when answering entirely randomly. With a .019728 chance of success, she has a 1 - .019728 = .980272 chance to fail. The chance to fail 771 times is...

.980272^771 = 2.12974*10^-7 = .00000021

Conclusively, there's a .000021% chance that when answering these questions completely randomly 771 times, she will have not passed even once.

SK.Testie

Canada11084 Posts

February 06 2009 09:54 GMT

#83

I'm gonna go with she's failing on purpose because she likes the people there and it's an excuse to get out of her house every day. =]

KOFgokuon

United States14888 Posts

February 06 2009 09:58 GMT

#84

On February 06 2009 13:09 KurtistheTurtle wrote:
aww poor old lady. She's really determined. If she hits 1k, just give it to her.

hell no get her off the road permanently

NathanSC

United States620 Posts

February 06 2009 10:05 GMT

#85

On February 05 2009 19:41 evanthebouncy! wrote:

Show nested quote +

I'll show some work. Say his average is 40%, how likely is she able to get more than 1 test over 60% in 800 tests?

Well, get more than 1 test in 800 tests is the same as saying
1 - P(getting all 800 under 60%)

Let us assume that each test is 10 problems
Let us assume that for each problem, she has .4 chance of answer correct, and .6 for incorrect
We want to answer: What's the probability that she passes score 6 with her chances?
Well, it's going to be
P(score6) + P(score7) + ... + P(10)
= .4^6*.6^4 + .4^7*.6^3 + ... + .4^10*.6^0
= 0.00138
Actually fairly slim chances.

Now getting all 800 under 60% that would be failing 800 times. Which is
(1 - 0.00138)^800 = 0.33

So that means she'll be passing with a probability of 0.67. Which isn't too bad lol.

The problem with your math is it discounts iterations entirely. This is very easy to notice if you try to calculate the chance of her to get any score using your formula:

.4^0*.6^10 + .4^1*.6^9 + ... + .4^10*.6^0
= .01793
= 1.793%

So you're essentially implying she has only a 1.793% chance of even getting a score. Period.

ZergZoul

Mexico408 Posts

February 10 2009 17:19 GMT

#86

Does someone else think she may be doing this on purpose ?
Like to get attention or something ?

Ryan307 :)

United States1289 Posts

February 10 2009 20:25 GMT

#87

On February 05 2009 18:55 Liquid`Drone wrote:
haha
"it was a record-breaking number here"
so someone else failed 770 times?

exactly what i thought lol

Racenilatr

United States2756 Posts

February 10 2009 20:32 GMT

#88

lol that is pretty sad, but when I think of my grandma, she fits that "stereotype" pretty well

-orb-

United States5770 Posts

February 10 2009 21:49 GMT

#89

Wowwwww what an idiot...

Instead of spending 3600 dollars on taking the test, spend a fucking hour studying the answers and then take the test once...

lxginverse

Monaco1506 Posts

February 23 2009 11:11 GMT

#90

its 775 times now according to Yahoo!

Too_MuchZerg

Finland2818 Posts

February 23 2009 11:13 GMT

#91

I am more scared if she passes the exam

D10

Brazil3409 Posts

February 23 2009 14:54 GMT

#92

Well, at least the test works, here in Brazil they would have said something like "hmmmm you passed but you need to gimme some money for administrative costs"

TimeShifter

Singapore235 Posts

February 23 2009 15:05 GMT

#93

On February 05 2009 13:05 baal wrote:

this is so freakin mean D:

KOFgokuon

United States14888 Posts

February 23 2009 22:37 GMT

#94

thanks for the update
...

freelander

Hungary4707 Posts

February 23 2009 22:42 GMT

#95

this thread should be updated with the actual new numbers

ZidaneTribal

United States2800 Posts

February 23 2009 22:44 GMT

#96

i hope she doesnt pass because she is gonna be a threat to everyone on the street

the day after she passes the test...

*skorea headline*

"old lady kills 20"