Probabilities with infinite sequences of letters A drunk poet writes an infinite sequence of letters, each one chosen uniformly at random from an alphabet of 26 letters, independently of the others.

*

*What is the probability that in the first $10$ digits there is at least one $A$?

My answer:
$$\frac{10\cdot 25^9}{26^{10}}$$


*What is the probability that in the first $10$ digits there are exactly $3$ A's?

My answer:
$$
\frac{25^7}{26^{10}}
$$
P.S. I do not if I must also multiply something depending on the positions assumed by the $A$'s, but I think not.


*What is the probability that in the first $11$ digits there are exactly $3$ A's and $2$ B's.

My answer:
$$
\frac{24^5}{26^{10}}
$$
P.S. I do not if I must also multiply something depending on the positions assumed by the $A$'s and the $B$'s, but I think not.


*Show that the probability that the drunk poet sooner or later writes the word PROBABILITY is $1$.

My answer: I have no clue.
 A: For the first two questions find $P(A\geq1)$ and $P(A=3)$ where $A$ is random variable having binomial distribution with parameters $n=10$ and $p=\frac1{26}$.
For the third question find $P(A=3,B=2,X=6)$ where $(A,B,X)$ have multinomial distribution with parameters $n=11$ and $(p_A,p_B,p_X)=(\frac1{26},\frac1{26},\frac{24}{26})$.
For the last question: Number the letters and for $n=1,2,\dots$ let $E_n$ be the event that the string "PROBABILITY" is not on the consecutive letters with numbers $100n,100n+1,\dots, 100n+10$.
Observe that these events are disjoint and independent and have equal probability.
If $E$ denotes the event that the string "PROBABILITY" does not occur in the infinite sequence for every $m$ we have:$$E\subseteq\bigcap_{n=1}^{m}E_n$$so that:$$P(E)\leq P\left(\bigcap_{n=1}^{m}E_n\right)=\prod_{n=1}^m P(E_n)=P(E_1)^m$$
Then on base of the facts that $P(E_1)<1$ and $m$ can be taken as large as we want we can conclude that $$P(E)=0\text{ or equivalently }P(E^c)=1$$
A: All your answers are wrong. Here are some hints

*

*You need to consider the cases where several A appear, which you did not. The easiest is to compute the probability of the opposite event, that no A appear.


*You need indeed to take into account that there are multiple ways of placing the 3 A into the 10 first letters


*Idem


*Have a look at the Borel-Cantelli lemma. This is one of the most famous application of this lemma
A: Saying "there is at least one A" is the same as saying "it is not the case that there are no A's". So find the probability that there is no A and subtract from 1.  The probability any specific letter is NOT an A is 25/26.  The probability that, in 10 letters, none is an A is (25/26)^10 so the probability there is at least one A is 1- (25/26)^10= (26^10- 25^10)/26^10.
To find the probability that there are exactly three A, first find the probability that the first three letters are As (1/26^3), then find the probability the last seven letters are NOT As (25^7/26^7), and multiply them (25^7/26^10).  Now multiply that by the number of different orders of 3 things of one kind (A) and 7 things of a different kind (not A) which is 3!7!
The probability of 3 As in 10 letters is 3!7!(25^7/26^10).
