Probability of taking the right key at the tenth pick

Question

There's 15 keys available and only one which is able to open the door. The keys are taken one by one without returning. Determine the probability of taking the right key at the tenth times of taking. 

My first thought is 1/6. Because on the 10th choice, it will be only 6 keys left with 1 right key... But Im not sure.

Please help me with clear step by step. Thx

Answer 1

Hint: Let $K_1,\ldots,K_{15}$ be the $15$ keys. Now suppose you select $10$ of the $15$ keys, where order matters. In how many ways can this be done? Call this number $N$.

Now suppose that the $10$th key was the correct key. Then the first $9$ keys must all have been incorrect. In how many ways could this happen? Call this number $M$.

Your answer will be $M/N$.

Answer 2

There are two possible readings of this problem:

1. Determine the probability that the right key is taken at the 10th pick.
2. Determine the probability that the right key is taken at the 10th pick, given that there actually is a tenth pick.

Your proposed answer is for the second question. It is correct; for, at the tenth pick, we are simply presented a collection of six keys, one of which is the right one.

However, I'm quite confident that the intended reading is the first one. This does not mean that our solution to problem 2 is worthless. It is a perfect fit for using conditional probability, the device for dealing with probabilities involving terms like "given".

If (for convenience of notation) we define the events $A$ and $B$ as follows:

\begin{align} A &= \text{The right key is taken at the tenth pick}\\ B &= \text{There is a tenth pick} \end{align}

then by definition of conditional probability:

$$\Pr(A \mid B) = \frac{\Pr(A \text{ and } B)}{\Pr(B)} \tag{1}$$

Now $A \text{ and } B$ is obviously just the same as $A$ (because we can only take the key at the tenth pick if there is a tenth pick). Also, our solution to problem 2. tells us that $\Pr(A \mid B) = \frac16$.

It remains to find $\Pr(B)$ -- after that, it is easy to find $\Pr(A)$ using the formula.

I will leave that to you, with the hint that:

There is a tenth pick precisely when the correct key is not among the first nine picks.

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Added (definitive solution): As noted in several comments, it's not hard to see that:

$$\Pr(\text{There is an $n+1$st pick} \mid \text{There is an $n$th pick}) = \frac{15-n}{16-n}$$

We can then prove by induction that:

$$\Pr(\text{There is an $n$th pick}) = \frac{16-n}{15} \tag{2}$$

For $n=1$, it's obviously true: There's always a first pick, so the probability is one. The inductive step easily follows from algebraic manipulation of the identity $(1)$ above (for appropriate choices of $A$ and $B$).

Giving $B$ its original meaning "There is a tenth pick" again, we can read $\Pr(B)$ from $(2)$: it's $\dfrac6{15}$.

In conclusion, then:

$$\Pr(\text{The right key is taken at the tenth pick}) = \frac16\frac{6}{15} = \frac1{15}$$

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Hmm. This result has a very simple form (this is often an indication that a more elegant argument exists). Indeed, this case is no different.

Obviously, there is no difference between picking a key every time, and deciding on an order for trying them beforehand. So we arrange the keys (denoted by $*$) like so:

$$*\ *\ *\ *\ *\ *\ *\ *\ *\ *\ *\ *\ *\ *\ *$$

Now, one of them is the correct key (let us indicate this by a box):

$$*\ *\ *\ *\ *\ *\ *\ *\ \boxed{*}\ *\ *\ *\ *\ *\ *$$

We're asking nothing but the probability that the box will be in tenth position. It is now extremely obvious that this probability is $\dfrac1{15}$ -- without all kinds of difficult calculations.

So, remember this: Sometimes looking at a problem from a different direction can make your life much easier. (And, when picking lottery tickets, it doesn't matter if you pick first or last: your probability of winning the prize doesn't change.)