# One woman with ten keys, one lock to be opened, and one heck of a question!

• Oct 19th 2011, 06:28 AM
One woman with ten keys, one lock to be opened, and one heck of a question!
The question:
A woman has 10 keys out of which only one opens a lock. She tries the keys one after another (keeping aside the failed ones) till she succeeds in opening the lock. What is the chance that it is the seventh key that works?

My approach so far:
From what I've thought out so far, I think we need to find out the probability of each of the keys failing before the seventh key, and then the probability of the seventh key being the one that works out of the remaining keys (which would be four by the time she reaches the seventh key).
The only problem I'm facing is that I don't understand how to take into account the probabilities of those events happening one after another, simultaneously.

• Oct 19th 2011, 06:41 AM
Plato
Re: One woman with ten keys, one lock to be opened, and one heck of a question!
Quote:

The question:
A woman has 10 keys out of which only one opens a lock. She tries the keys one after another (keeping aside the failed ones) till she succeeds in opening the lock. What is the chance that it is the seventh key that works?

QUESTION
Once she tries a key and it fails, what does she with it?
Does she discord it? Or put is back in the pile?
• Oct 19th 2011, 06:58 AM
Re: One woman with ten keys, one lock to be opened, and one heck of a question!
so the next probability that I'l have to evaluate will be from one less number of keys.
So I think I'd do it this way:
The probability of the first key failing is 9/10 ('cause out of ten, any nine can fail).
Then she keeps it aside.
So the probability of the next key failing from the keys remaining will be 8/9 ('cause out of the nine keys left, only one will work).
And so on up to 4/5, after which she'll set aside the failed key and reach the seventh key, and have four keys left, out of which only one will work. So, the probability of picking up that key will be 1/4.

What I don't understand is how to find the probability of these events happening together.
• Oct 19th 2011, 07:02 AM
Soroban
Re: One woman with ten keys, one lock to be opened, and one heck of a question!

Quote:

A woman has 10 keys out of which only one opens a lock.
She tries the keys one after another (keeping aside the failed ones)
till she succeeds in opening the lock.
What is the probability that it is the seventh key that works?

The problem is much simpler than you think.

Place the ten keys in a row in random order.

Consider the 3rd key.
What is the probability that it is the correct key?
. . Answer: .$\displaystyle \tfrac{1}{10}$ . . . agree?

Consider the 9th key.
What is the probability that is the correct key?
. . Answer: .$\displaystyle \tfrac{1}{10}$ . . . right?

Now consider the 7th key . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can do it your way.

If the 7th key is correct, the first six are incorrect.
. . (and we don't care about the last three.)

So, we want: . no - no - no - no - no - no - yes.

The probability is: .$\displaystyle \frac{9}{10}\cdot\frac{8}{9}\cdot\frac{7}{8}\cdot \frac{6}{7}\cdot\frac{5}{6}\cdot \frac{1}{5}$

. . which reduces to: .$\displaystyle \frac{\rlap{/}9}{10}\cdot\frac{\rlap{/}8}{\rlap{/}9}\cdot\frac{\rlap{/}7}{\rlap{/}8}\cdot \frac{\rlap{/}6}{\rlap{/}7}\cdot\frac{\rlap{/}5}{\rlap{/}6}\cdot \frac{1}{\rlap{/}5} \;=\;\frac{1}{10}$

• Oct 19th 2011, 07:06 AM
Plato
Re: One woman with ten keys, one lock to be opened, and one heck of a question!
Quote:

so the next probability that I'l have to evaluate will be from one less number of keys.

Then this problem is no different by symmetry than asking for the probability that the second key works:
$\displaystyle \frac{9}{10}\cdot\frac{1}{9}=\frac{1}{10}$.
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