Question

 A woman has 10 keys 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 opens the lock?
[a] $\dfrac{7}{10}$
[b] $\dfrac{1}{2}$
[c] $\dfrac{3}{10}$
[d] $\dfrac{1}{10}$

Answer 1

Hint: Use the fact that $P\left( A\bigcap B \right)=P\left( A \right)\times P\left( B|A \right)$. Assume that ${{E}_{i}}$ is the event that the ${{i}^{th}}$ key does not open the lock. Using the above property, find the probability that the seventh key opens the lock. Alternatively, find the total possible combinations of the keys in which the correct is placed at the seventh position and the total possible combinations of the keys.

Complete step by step answer:
Let ${{E}_{i}}$ be the event that the ${{i}^{th}}$ key does not open the lock.
Hence, we have to find the probability of the event ${{E}_{1}}\bigcap {{E}_{2}}\bigcap \cdots \bigcap {{E}_{6}}\bigcap \left( {{E}_{7}} \right)'$
Now, we know that
$P\left( {{E}_{1}}\bigcap {{E}_{2}}\bigcap \cdots \bigcap {{E}_{6}}\bigcap \left( {{E}_{7}} \right)' \right)=P\left( {{E}_{1}} \right)\times P\left( {{E}_{2}}|{{E}_{1}} \right)\times P\left( {{E}_{3}}|{{E}_{1}}\bigcap {{E}_{2}} \right)\cdots \times P\left( {{E}_{6}}|\bigcap\limits_{i=1}^{5}{{{E}_{i}}} \right)\times P\left( \left( {{E}_{7}} \right)'|\bigcap\limits_{i=1}^{6}{{{E}_{i}}} \right)$
Now, we have
$P\left( {{E}_{1}} \right)=\dfrac{9}{10}$(Since 9 keys cannot open the lock (Favourable cases) and there are a total 10 keys (Total cases)).
$P\left( {{E}_{2}}|{{E}_{1}} \right)=\dfrac{8}{9}$(Since now there are 8 keys which cannot open the lock (Favourable cases) and there are a total of 9 keys (Total keys))
Continuing this way, we get
$P\left( {{E}_{3}}|{{E}_{1}}\bigcap {{E}_{2}} \right)=\dfrac{7}{8},\cdots ,P\left( {{E}_{6}}|\bigcap\limits_{j=1}^{5}{{{E}_{i}}} \right)=\dfrac{4}{5}$
Also, we have
$P\left( \left( {{E}_{7}} \right)'|\bigcap\limits_{j=1}^{6}{{{E}_{i}}} \right)=\dfrac{1}{4}$ (Since there is only one key that can open the lock and there are a total of 4 keys left).
Hence, we have
$\begin{align}
  & P\left( {{E}_{1}}\bigcap {{E}_{2}}\bigcap \cdots \bigcap {{E}_{6}}\bigcap \left( {{E}_{7}} \right)' \right)=\dfrac{9}{10}\times \dfrac{8}{9}\times \dfrac{7}{8}\times \cdots \times \dfrac{4}{5}\times \dfrac{1}{4} \\
 & =\dfrac{1}{10} \\
\end{align}$
Hence the probability that the seventh key opens the lock is $\dfrac{1}{10}$
Hence option [d] is correct.

Note: Alternative solution:
The number of ways in which correct key is at seventh positions is equal to the number of ways in which we can place the correct key at seventh position and arrange the remaining 9 keys in 9 places, which can be done in $1\times 9!$
Also, the total number of ways in which the keys can be placed is $10!$
Hence the probability that the seventh key is the correct keys is $\dfrac{9!}{10!}=\dfrac{1}{10}$, which is the same as obtained above.
Hence option [d] is correct.