Question

Five letters are placed at random in five addressed envelopes. The probability that all the letters are not dispatched in the respective right envelopes is

• A. \(\frac {4}{5}\)
• B. \(\frac {119}{120}\)
• C. \(\frac {1}{120}\)
• D. \(\frac {1}{5}\)

Answer 1

The Correct Option is B

Let's consider the first letter. The probability that it is not placed in the correct envelope is \(\frac {4}{5}\) since there are 4 envelopes remaining and only 1 of them is the correct envelope. 
Now, let's move on to the second letter. The probability that it is not placed in the correct envelope depends on two cases: either the first letter was placed in the correct envelope, or it was not. 
Case 1: The first letter was placed in the correct envelope. In this case, the second letter has 4 envelopes remaining, and there is only 1 correct envelope for it. So the probability of the second letter not being placed in the correct envelope is \(\frac {1}{4}\). 
Case 2: The first letter was not placed in the correct envelope. In this case, the second letter has 4 envelopes remaining, but none of them is the correct envelope for it. So the probability of the second letter not being placed in the correct envelope is \(\frac {4}{4}\) = 1. 
Since these two cases are mutually exclusive (they cannot happen at the same time), we can add their probabilities: 
Probability of the second letter not being placed correctly = \(\frac {1}{5}\) x \(\frac {1}{4}\) + \(\frac {4}{5}\) x (1) = \(\frac {1}{20}\) + \(\frac {4}{5}\) = \(\frac {9}{20}\). 
We can continue this process for the remaining letters: 
Probability of the third letter not being placed correctly = \(\frac {1}{5}\) x \(\frac {1}{3}\) + \(\frac {4}{5}\) x \(\frac {9}{20}\) = \(\frac {1}{15}\) + \(\frac {36}{100}\) = \(\frac {21}{100}\). 
Probability of the fourth letter not being placed correctly = \(\frac {1}{5}\) x \(\frac {1}{2}\) + \(\frac {4}{5}\) x \(\frac {21}{100}\) = \(\frac {1}{10}\) + \(\frac {84}{100}\) = \(\frac {19}{50}\) 
Probability of the fifth letter not being placed correctly = \(\frac {1}{5}\) x \(\frac {1}{1}\) + \(\frac {4}{5}\) x \(\frac {19}{50}\) = \(\frac {1}{5}\) + \(\frac {1}{5}\) = \(\frac {11}{25}\). 
Now, we multiply these probabilities together to find the probability that all the letters are not dispatched in the respective right envelopes: 
Probability = \(\frac {4}{5}\) x \(\frac {9}{20}\) x \(\frac {21}{100}\) x \(\frac {19}{50}\) x \(\frac {11}{25}\) = \(\frac {119}{120}\). 
Therefore, the correct answer is option (B) \(\frac {119}{120}\).