MadeEasy Workbook: Probability

Question

Q : There are two identical locks with two identical keys and the key are among the six different ones which a person carries in his pocket.In hurry he drops one key somewhere . Then the probability that the locks can still be opened by drawing one key at random is equal to :

a) 1/3

b) 5/6

c) 1/12

d) 1/30

Comments

Iam getting $\frac{1}{3}$

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Plz explain how

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yes 1/3

Answer 1

In above problem, we have $6-keys$ out of which $2$ are identical and any of these $2$ can open lock.

One key is dropped. So, the dropped key may be the one that opens lock with a probability of $\frac{2}{6}$ and may be the one that doesn't opens door with probability of $\frac{4}{6}$. (here the problem is divided into $2$ cases)

Case 1: Key that opens door is lost

Now, we are remaining with $5$ keys out of which only one opens the lock.

Case 2: Key that doesn't opens door is lost

Now, we are remaining with $5$ keys out of which $2-keys$ open the lock.

P(lock-opens) = Key that opens door is lost and lock opens (or) Key that doesn't opens door is lost and lock opens

$P(lock-opens) = \frac{2}{6}*\frac{1}{5}  +  \frac{4}{6}*\frac{2}{5} = \frac{10}{30} = \frac{1}{3}$

Comments

Thanks....I got it.Nice explanation.

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my pleasure.

Answer 2

Here, 6 keys in his pocket.

Among them 5 are remaining

So, probability of searching key amongremaining= $\frac{5}{6}$

Now, we want to save both the keys which opens lock.

So, probability of getting that= $\frac{2}{5}$

So, total probability= $\frac{5}{6}\times \frac{2}{5}$= $\frac{1}{3}$