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The key to opening is in a bunch of ten keys. A man attempts to open the door by trying the keys at random and discarding the wrong key. The probability that door is opened on fifth trial is:

The solution key involves using multiplication rule of probability for dependent events:

$$ P = \frac{9}{10} \frac{8}{9} \frac{7}{8} \frac{6}{8} \frac{1}{6} = \frac{1}{10}$$

Explanation, nine out of the ten are wrong and hence the probability of choosing wrong key in first choose must be $ \frac{9}{10}$, similarly once we have taken one wrong key , the probability of choosing the next key is given as $ \frac{8}{9}$ and so on till the fifth draw where we only have one 'right' key out of the six.

However, if we were to extend this result, it'd turn out that the chance of man opening the door on first try, second try, third try.. last try are all equal.. how is this possible?

I think that if we do a truly random experiment, then if we get wrong answer the first few times, it must increase our chances of getting the right answer later on.

tryst with freedom
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    Of course the probability increases as you discard keys, but this is offset by the fact that it is unlikely that you'll have a long string of wrong keys. Just think of it this way: suppose all the keys are arranged randomly in a line. Then the winning key is equally likely ot be in any position in that line. – lulu Feb 08 '21 at 13:41

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The probability that the door is opened on the first, second,....tenth trial is always constant and equal to $\frac{1}{10}$

The probability cannot change...

think at the following problem: you have only 1 good key among 10. the probablity to get it cannot depend by the fact that it is detected at first, second or tenth draw

tommik
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