There are two identical locks with two identical keys and the key are among six different ones which a person carries in his pocket. In a hurry, he drops one key somewhere. Then the probability that the locks can still be opened by drawing one key at random is?
I am a newbie to probability and I am learning it for my competitive exam, so please forgive me if I ask some really silly questions.
Here if I understand the problem, then there are 2 similar locks and they have the same key,(Means we have 2 keys and either of them is capable of opening the lock). Now, these two keys are in a set of 6 keys. One key is dropped by the person.
The probability that locks can still be opened by drawing one key at random should be like:
The favourable case will be when the keys of locks are still available=2.
Total cases=5(One key is lost)
So, probability=$\frac{2}{5}$
Am I correct?
Best Answer
Your answer is not correct.
Given that there are $6$ keys in which $2$ are identical and with $2$ keys we can open the lock.
Given that he dropped $1$ key. So, $5$ are remaining.
So, the probability is $\dfrac56$
Now, we need both keys which can open the lock.
So, the probability is $\dfrac25$
Now, the total probability is $\dfrac56\times\dfrac25=\dfrac13$