Question

There are 2 locks on the door and the keys are among the six different ones you carry in your pocket. In a hurry you dropped one somewhere. The probability that you can still open the door is
a. $\dfrac{1}{2}$
b. $\dfrac{1}{3}$
c. $\dfrac{2}{3}$
d. $\dfrac{1}{4}$

Answer 1

Hint: We will first find the probability that the door will not be opened and then we will subtract the obtained probability from 1, to get the probability that we can still open the door. We will use the formula of probability, that is, $\text{probability=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$.

Complete step-by-step answer:
It is given in the question that there are 2 locks on the door and the keys are among the six different ones we carry in our pocket. In a hurry we dropped one somewhere. And we have been asked to find the probability that we can still open the door.
So, we have 2 keys out of 6 keys with which we can open the door. Let us consider the first key as A and the second key as B.
Now, we will find the probability that the key lost will be A. So, we know that out of the 6 keys, 1 key is lost, which is A. So, we get the total number of outcomes as 6 and the number of favourable outcomes of not opening the door as 1. So, on putting these values in the formula of probability, that is, $\text{probability=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$, we get $P\left( A \right)=\dfrac{1}{6}$.
Similarly, we will find the probability that the lost key is B. So, out of the 6 keys, 1 key, that is key B will be lost. So, we get the total number of outcomes as 6 and the number of favourable outcomes of not not opening the door as 1. So, on putting these values in the formula of probability, that is, $\text{probability=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$, we get $P\left( B \right)=\dfrac{1}{6}$.
Now, we will find the probability that we cannot open the door. So, we will not be able to open the door if the key is lost. Therefore, we can find the probability that we cannot open the door as,
(Probability of losing key A) + (Probability of losing key B)
P(A) + P(B)
$\begin{aligned} & =\dfrac{1}{6}+\dfrac{1}{6} \\\ & =\dfrac{2}{6} \\\ \end{aligned}$
Hence, the probability that we cannot open the door will be $\dfrac{2}{6}$.
We know that the total probability of anything will be equal to 1. It means that, sum of the probability that we can open the door and we cannot open the door will be equal to 1. So, we can write,
(Probability that we can open the door) + (Probability that we cannot open the door) = 1
We have already found that the probability that we cannot open the door is $\dfrac{2}{6}$. So, on substituting this value in the above expression, we will get,
(Probability that we can open the door) + $\dfrac{2}{6}$ = 1
(Probability that we can open the door) = $1-\dfrac{2}{6}$
$\begin{aligned} & =\dfrac{6-2}{6} \\\ & =\dfrac{4}{6} \\\ & =\dfrac{2}{3} \\\ \end{aligned}$
Thus, we get the probability that we can open the door as $\dfrac{2}{3}$.
Thus option (c) is the correct answer.

Note: The possible mistake that the students can make is by directly solving this question, they may solve it, thinking that 1 key is lost out of 6 keys, so the probability that we can still open the door will be $\dfrac{1}{6}$, but this is wrong because there are two locks in the door and we cannot assume that the lost key is the correct key that can open the lock.