SOLUTION: Suppose you have a key ring with eight keys on it, one of which is your house key. Further suppose you get home after dark and cant see the keys on the key ring. You randomly try o

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose you have a key ring with eight keys on it, one of which is your house key. Further suppose you get home after dark and cant see the keys on the key ring. You randomly try o      Log On


   

Question 957876: Suppose you have a key ring with eight keys on it, one of which is your house key. Further suppose you get home after dark and cant see the keys on the key ring. You randomly try one key at a time, being careful not to mix the keys you have already tried with the ones you have not. What is the probability that you get the right key on the fifth try.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the probability of getting the key right on any one try is 1/8.
the probability of getting the key wrong on any one try is 7/8.

in order to get the key on the fifth try, you have to have failed on tries 1 to 4.

once you get the key you can stop.

the probability of failing on the first 4 tries and getting it on the fifth try is equal to 7/8 * 6/7 * 5/6 * 4/5 * 1/4.

the probability is therefore (7 * 6 * 5 * 4 * 1) / (8 * 7 * 6 * 5 * 4) which is equal to 1/8.

the probability is the same whether you are looking for the probability of getting it on the first try or on the second try or on the third try, etc.

first try is 1/8

second try is 7/8 * 1/7 = (7 * 1) / (8 * 7) = 1/8

third try is 7/8 * 6/7 * 1/6 = (7 * 6 * 1) / (8 * 7 * 6) = 1/8.

suppose you have 1 out of 4.

the probability of getting it on any try is 1/4.

suppose you want to determine the probability of getting it on the 3d try.

suppose c is the right key.
a and b and d are not the right key.

all possible permutations of 4 keys are:

abcd ***
abdc
acbd
acdb
adbc
adcb ***

bacd ***
badc
bcad
bcda
bdac
bdca ***
cabd
cadb
cbad
cbda
cdab
cdba

dabc
dacb ***
dbac
dbca ***
dcab
dcba

24 possible permutations in all.

out of these 24, c is in the third position 6 out of the 24 possible permutations.

that's a probability of 6/24 = 1/4.

look for the probability of getting key c on the fourth try and the probability should be the same.

all possible permutations of 4 keys are:

abcd
abdc ***
acbd
acdb
adbc ***
adcb

bacd
badc ***
bcad
bcda
bdac ***
bdca
cabd
cadb
cbad
cbda
cdab
cdba

dabc ***
dacb
dbac ***
dbca
dcab
dcba

probability is still 6/24 = 1/4.