Answer:
[tex]\frac{1}{6}[/tex]
Step-by-step explanation:
There are 3 cards which are to be placed in 3 envelopes. We need to find the probability of each card being placed in its correct envelope. For this first we need to find out to total possible number of arrangements of adding 3 cards in the 3 envelopes. Since the order of placing the cards in the envelope matters here, this is a problem of permutations. So we have to find the permutations of 3 objects taken 3 at a time which is represented by 3P3.
The formula for permutations is:
[tex]^{n}P{r}=\frac{n!}{(n-r)!}[/tex]
So, for the given case, it would be:
[tex]^{3}P_{3}=\frac{3!}{(3-3)!}=6[/tex]
This means there are total 6 possible ways of adding the cards to the envelopes. But only 1 correct way of placing the cards in the correct envelopes.
So, the probability that all three cards were placed in the correct envelope = [tex]\frac{1}{6}[/tex]