Question

The key for opening a door is in a bunch of 10 keys. A man attempts to open the door by trying the keys at random, discarding the wrong key. The probability that the door opens in the fifth trial is?

Answer 1

Hint: We will individually find the probability of each key noticing that we need to get 4 keys wrong and then fifth one as the correct key. Now, multiply them all to get the answer.

Complete step-by-step answer:
Since, we have the key for door in a bunch of 10 keys. There must be 9 wrong keys and 1 right key in those 10 keys. Now, since we need the right key in fifth trial, therefore, we require 4 wrong keys at first and then the right key.
Since, we have 9 wrong keys, so while selecting first key, we can select any 9 among total 10 keys.
Therefore, P (selecting the first key) = $\dfrac{9}{{10}}$
Now, first key has gone and we are left with 8 wrong keys and total 9 keys.
Now, we need to select second key to be wrong as well.
Therefore, P (selecting the second key) = $\dfrac{8}{9}$
Now, second key has gone and we are left with 7 wrong keys and total 8 keys.
Now, we need to select third key to be wrong as well.
Therefore, P (selecting the third key) = $\dfrac{7}{8}$
Now, third key has gone and we are left with 6 wrong keys and total 7 keys.
Now, we need to select the fourth key to be wrong as well.
Therefore, P (selecting the fourth key) = $\dfrac{6}{7}$
Now, fourth key has gone and we are left with 5 wrong keys and a total 6 keys.
Now, we need to select the fifth key to be the right one.
Therefore, P (selecting the fifth key) = $\dfrac{1}{6}$
Now, the probability of getting the right key in fifth trial = $\dfrac{9}{{10}} \times \dfrac{8}{9} \times \dfrac{7}{8} \times \dfrac{6}{7} \times \dfrac{1}{6}$.

Simplifying the calculations on the right hand side, we will get the probability of getting the right key in the fifth trial is $\dfrac{1}{{60}}$.

Note:
The students must note that we multiplied all the probabilities because they are independent of each other and when we find the probability of intersection of events, we multiply them, if they are independent.
The students must note that the probability of any event E is given by $P(E) = \dfrac{{n(F)}}{{n(S)}}$