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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.

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

My answer:

$$\frac{10\cdot 25^9}{26^{10}}$$

  1. 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.

  1. 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.

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

My answer: I have no clue.

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3 Answers 3

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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 $$P(E)=0\text{ or equivalently }P(E^c)=1$$

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All your answers are wrong. Here are some hints

  1. 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.

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

  3. Idem

  4. Have a look at the Borel-Cantelli lemma. This is one of the most famous application of this lemma

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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).

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    $\begingroup$ I do not understand why I have to multiply by 3!7!Can you explain me better, please? Moreover, what about question 3? $\endgroup$ Aug 16 at 13:07
  • $\begingroup$ It must not be multiplied by $3!7!$ but by $\binom{10}3=\frac{10!}{3!7!}$ $\endgroup$
    – drhab
    Aug 16 at 14:23
  • $\begingroup$ Yes, I miswrote. Thans for the correction. $\endgroup$ Aug 16 at 16:11

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