Question

The key to 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 is opened in the fifth trial is.

Answer 1

Hint: To solve the given question, we will first find out what is the probability that the first key drawn is wrong. Now, we will find out of the keys left what is the probability that the key drawn is wrong. We will do this four times in a row removing the key from the bunch each time. After this, we will find out what is the probability that out of the keys left, the key drawn is correct. Now, we will multiply all these probabilities to get the answer.

Complete step by step answer:
We are given the question that there are 10 keys in a bunch. After testing the keys, they are removed from the bunch if they are wrong and the random keys are drawn from the remaining bunch of keys. Now, we know that the probability of any event is denoted by P(E) and it is obtained by dividing the favorable outcomes with the total number of outcomes, so, we can say that,
$P\left( E \right)=\dfrac{\text{Favorable Outcomes}}{\text{Total Number of Outcomes}}$
Now, in the first draw, we have to select the wrong key. In the bunch, there are 9 wrong keys and 1 correct key. So, the probability that a wrong key is selected in the first draw in given by
$P\left( \text{First draw} \right)=\dfrac{9}{10}$
Now, the wrong key which we have selected in the first draw has been removed from the bunch. Now, there are 9 keys left out of which 8 are wrong and one is correct. The probability that the key is drawn this time (second time) is wrong is given by,
$P\left( \text{Second draw} \right)=\dfrac{8}{9}$
Now, in the third draw also, the key drawn is incorrect. There are 8 keys left out of which 7 are wrong. The probability of happening this is,
$P\left( \text{Third draw} \right)=\dfrac{7}{8}$
Similarly, we can say that the probability of drawing the wrong key is the fourth draw is,
$P\left( \text{Fourth draw} \right)=\dfrac{6}{7}$
Now, in the fifth draw, the man draws the correct key out of the six keys left in the bunch. The probability of doing this is,
$P\left( \text{Fifth draw} \right)=\dfrac{1}{6}$
Now, the probability that the key is opened in the fifth trial will be obtained by multiplying the probabilities of the first, second, third, fourth, and fifth draw.
Probability = P (First draw) . P (Second draw) . P (Third draw) . P (Fourth draw) . P (Fifth draw)
$\Rightarrow \text{Probability}=\dfrac{9}{10}\times \dfrac{8}{9}\times \dfrac{7}{8}\times \dfrac{6}{7}\times \dfrac{1}{6}$
$\Rightarrow \text{Probability}=\dfrac{1}{10}$
Thus the probability that the correct key will be drawn on the fifth trial is $\dfrac{1}{10}.$

Note: We can also say directly that the probability of the key being correct in the fifth trial is $\dfrac{1}{10}.$ This is because here the probability does not depend on which trial we are getting the correct key. The probability for the key being correct in any trial will be $\dfrac{1}{10}$ i.e,
$P=\dfrac{\text{Favorable Outcomes}}{\text{Total Number of Outcomes}}=\dfrac{1}{10}$