Question

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

Answer 1

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

Answer 2

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

Answer 3

the chances are:

P(1st) = 1/7
P(2nd) = 1/6
P(3rd) = 1/5

that aint it, is it

Answer 4

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

feels better

Answer 5

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...

Answer 6

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...

Answer 7

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...

Answer 8

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

Answer 9

\[\frac{^NC_m\ ^?C_?}{^NC_x}\]

cant recall the formula or even if its right

Answer 10

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

Answer 11

Zarkon is keen with these

Answer 12

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

Answer 13

Zarkon, can you explain that please?

Answer 14

And is vitor's explanation correct?

Answer 15

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

Answer 16

i was thinking about the hypergeometric probability distribution

Answer 17

this is a discrete uniform distribution

Answer 18

not that mine is correct, but this would explain what i was thinking better

http://129.81.170.14/~slukens/Math114_fall2009/Hypergeometric.pdf

Answer 19

How does Vitor's explanation differ from yours Zarkon?

Answer 20

Vitors is not in function format ...

Answer 21

Oh righttt nevermind

Answer 22

Got it...thanks all for the help!

Answer 23

you got the answer? or got the explanation :)

Answer 24

casue id love to know the answer lol

Answer 25

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!