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

Answer: (B)

$\frac {119}{120}$

Answer 2

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) $\frac {119}{120}$.