The probability of inviting three friends on 5 consecutive days, exactly one friend a day and no friend is invited on more than two days is

There are 3 friends and 5 days. Each day, one friend is invited, so there are $3^5 = 243$ total ways to invite friends.

Let the friends be $A, B, C$. We need to count the number of ways such that each friend is invited at most twice. Since there are 5 days, the possible distributions of the number of days each friend is invited are $(2, 2, 1)$ in some order.

The number of ways to choose which friend is invited once is $^{3}C_{1} = 3$.

For a specific distribution, say $A$ is invited 1 time, $B$ is invited 2 times, and $C$ is invited 2 times, the number of ways to arrange these invitations over 5 days is the number of permutations of the multiset $\{A, B, B, C, C\}$, which is $\dfrac{5!}{1!2!2!} = \dfrac{120}{4} = 30$.

Since there are 3 choices for the friend invited once, the total number of favorable outcomes is $3 \times 30 = 90$.

The probability is $\dfrac{90}{243}$.

Dividing both numerator and denominator by 9, we get $\dfrac{10}{27}$.

Answer: $\dfrac{10}{27}$