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Mathematics 83 Online
OpenStudy (anonymous):

Help with a probability function question! A woman has 7 keys on a keyring, one of which fits the door she wants to unlock. She randomly selects a key and tries it. If it does not unlock the door, she randomly selects another from these remaining and tries to unlock the door with it She continues in this manner until the door is unlocked. Let X be the number of keys she tries before unlocking the door, counting the key that actually worked. Find the probability function of X. Help with a probability function question! A woman has 7 keys on a keyring, one of which fits the door she wants to unlock. She randomly selects a key and tries it. If it does not unlock the door, she randomly selects another from these remaining and tries to unlock the door with it She continues in this manner until the door is unlocked. Let X be the number of keys she tries before unlocking the door, counting the key that actually worked. Find the probability function of X. @Mathematics

OpenStudy (amistre64):

\[^7C_x\ (\frac{1}{7})^x(\frac{6}{7})^{1-x}\] maybe

OpenStudy (amistre64):

that wont do, you cant have 6 keys fit ...

OpenStudy (amistre64):

the chances are: P(1st) = 1/7 P(2nd) = 1/6 P(3rd) = 1/5 that aint it, is it

OpenStudy (amistre64):

P(y) = 1/7 P(n,y) = 6/7 * 1/6 P(n,n,y) = 6/7 * 5/6 * 1/5 feels better

OpenStudy (anonymous):

The probability of opening in the first key is 1/7 The probability of opening in the second key is the probability of NOT opening on the first one * the probability of opening if she started with 6 keys, so 6/7 * 1/6 Then, not opening on the first, not opening on the second and opening if she had 5 keys 6/7*5/6*1/5 and it goes on...

OpenStudy (anonymous):

I understand how to come up with that but Im not sure how I would come up with the function...that's the part where Im not sure how to approach...

OpenStudy (anonymous):

The probability of opening in the first key is 1/7 The probability of opening in the second key is the probability of NOT opening on the first one * the probability of opening if she started with 6 keys, so 6/7 * 1/6 Then, not opening on the first, not opening on the second and opening if she had 5 keys 6/7*5/6*1/5 and it goes on...

OpenStudy (anonymous):

1/7 = 1/7 6/7 * 1/6 = 1/7 6/7 * 5/6 * 1/5 = 1/7 ...

OpenStudy (amistre64):

\[\frac{^NC_m\ ^?C_?}{^NC_x}\] cant recall the formula or even if its right

OpenStudy (anonymous):

That formula doesnt make sense haha...Im still trying to figure it out...thanks everyone for the help though...very useful!

OpenStudy (amistre64):

Zarkon is keen with these

OpenStudy (zarkon):

\[f_X(x)=\left\{\begin{matrix}\frac{1}{7} &\text{if} &x\in\{1,2,3,4,5,6,7\} \\0 && \text{otherwise}\end{matrix}\right.\]

OpenStudy (anonymous):

Zarkon, can you explain that please?

OpenStudy (anonymous):

And is vitor's explanation correct?

OpenStudy (zarkon):

what is there to explain. if x takes on any of the (integer) values 1 through 7 then P(X=x)=1/7 otherwise the answer is 0

OpenStudy (amistre64):

i was thinking about the hypergeometric probability distribution

OpenStudy (zarkon):

this is a discrete uniform distribution

OpenStudy (amistre64):

not that mine is correct, but this would explain what i was thinking better http://129.81.170.14/~slukens/Math114_fall2009/Hypergeometric.pdf

OpenStudy (anonymous):

How does Vitor's explanation differ from yours Zarkon?

OpenStudy (amistre64):

Vitors is not in function format ...

OpenStudy (anonymous):

Oh righttt nevermind

OpenStudy (anonymous):

Got it...thanks all for the help!

OpenStudy (amistre64):

you got the answer? or got the explanation :)

OpenStudy (amistre64):

casue id love to know the answer lol

OpenStudy (anonymous):

I understand how vitor's and zarkon's answers relate: The first key does have a probability of 1/7 The prob of not getting the key on first try and then getting it on the second try is: (6/7)*(1/6) = 1/7 The prob of not getting key on first try, not on second try, but on third try is: (6/7)*(5/6)*(1/5) = 1/7 And so on...all of them are equal to probability of 1/7. It is a 0 probability otherwise!

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