key ring

Mar 2010
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Jan gets home late, it's dark and she cannot see her keys. Her key ring has 5 similar keys, so she is going to pick a key to try to open the door. Two of the keys will open the door and the other keys will not. What is the probability she will get the door open on the first or second try?


so does this mean she has a 2/5 chance of 1st or 2nd try, or does this not include every possibility

i also thought it could be 4/25, but probability is brand new to me and i am not sure which way would be correct???????????
 

Plato

MHF Helper
Aug 2006
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She has a probability of \(\displaystyle \frac{2}{5}\) opening on the first try.
The a probability of \(\displaystyle \left(\frac{3}{5}\right)\left(\frac{2}{5}\right)\) of opening it on the second try. WHY?
SO WHAT?
 
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Mar 2010
110
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um..... she tries a key on the first try and if it opens it great, but if not, then wouldnt she have a 2/4 chance of opening it since the one key on the first try did not work?????
 
Mar 2010
110
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this would give her a (2/5) * (2/4) chance...is my thinking not right here?
1st 2nd
Plato,
I understand:
She has a probability of opening on the first try.

I do not understand why she has
The a probability of of opening it on the second try. WHY?
 

Plato

MHF Helper
Aug 2006
22,455
8,631
this would give her a (2/5) * (2/4) chance...is my thinking not right here?
1st 2nd
Plato,
I understand:
She has a probability of opening on the first try.

I do not understand why she has
The a probability of of opening it on the second try. WHY?
Well it depends upon how one reads the problem.
It says that she cannot see the keys.
So I assume that she could try the same key again.
That may not be a correct understanding.
You seem to read it as if she would not try the same key twice.
I don't which is correct.
 
Mar 2010
110
0
Okay, so first, let us say he cannot see the keys, then he would have a 2/5 chance on the first try and a 2/5 chance on the second try for a probability of 4/25...correct?

why did you say that he had a 2/5 chance on the first then a (3/5)(2/5) chance on the second try...this i do not understand?


Then, second, if he keeps the first key he tries separated from the other keys, he would have a 2/5 chance on the first try and a 2/4 chance on the second try for a probability of 4/20...correct?

sorry, this stuff is new to me, thanks for helping
 

Plato

MHF Helper
Aug 2006
22,455
8,631
why did you say that he had a 2/5 chance on the first then a (3/5)(2/5) chance on the second try...this i do not understand?
Because he must have failed on the first try.
The probability of failure is (3/5).
The probability of first success on the third try equals \(\displaystyle \left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)\)
He fails twice and then succeeds.
 
Mar 2010
110
0
Plato:
The question to answer is: What is the probability she will get it open on the first or second try?

Okay, so first, let us say he cannot see the keys, then he would have a 2/5 chance on the first try and a 2/5 chance on the second try for a probability of 4/25...correct?

Then, second, if he keeps the first key he tries separated from the other keys, he would have a 2/5 chance on the first try and a 2/4 chance on the second try for a probability of 4/20...correct?


So is my thinking way off?
 

Plato

MHF Helper
Aug 2006
22,455
8,631
This is my final reply. Assume that she cannot see the keys.
The probability that she opens the door on her first or second try is:
\(\displaystyle \frac{2}{5}+\left(\frac{3}{5}\right) \left(\frac{2}{5}\right) \).

Now assume that she can see the keys and can hold one back.
The probability that she opens the door on her first or second try is:
\(\displaystyle \frac{2}{5}+\left(\frac{3}{5}\right) \left(\frac{2}{4}\right) \).
 
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