A number of n digits is randomly generated. Find the probability that the number is a numerical palindrome.

How do you denote the sample space of this problem?

My try.

$\Omega = \{a_1a_2a_3...a_n\} $

a) I wasn't sure whether to write it with commas or not. I mean $\Omega = \{a_1,a_2,a_3...,a_n\}$

The event space, $F = P(\Omega)$, where P denotes the power set. And its cardinality $$|P(\Omega)|=2^{10^n}$$

  • $\begingroup$ The sample space is the number of $b$-long sequences. It's probably better to write then as strings of digits. WIth commas you'd want $\{(a_1, \ldots , a_n)\}. The events are sets of sequences, as you say. $\endgroup$ – Ethan Bolker May 6 at 20:37
  • $\begingroup$ And how do you denote $10^n$ strings of digits? $\endgroup$ – roy212 May 6 at 20:51

The sample space is defined as the set of all possible outcomes. Since it is a set, it is better to write it with commas using the standard $\{a, b, c, \dots\}$ notation.

But, in fact, your sample space $\Omega$ is actually the set of all possible $n$-digit numbers, rather than the set of digits of a given number. I assume this is what you had meant by $\{a_1, a_2, a_3, \dots, a_n\}$.

Depending on whether or not you include zeros at the beginning of the number, your sample space will be a different size, but one way of denoting a set of $n$-digit numbers would be to consider them to be $n$-length strings. The $a$th string could be denoted as $d_{a_1} d_{a_2} d_{a_3} \dots d_{a_n}$ and your sample space would look like $$\Omega = \{d_{1_1} d_{1_2} \dots d_{1_n}, d_{2_1} d_{2_2} \dots d_{2_n}, \dots, d_{m_1} d_{m_2} \dots d_{m_n}\}$$

where there are $m$ possible $n$-length strings. Note that $m = |P(\Omega)|$, as I think you know. This is equal to $10^n$ as you suggested, if you include leading zeros.

The set of events is actually not equal to the power set of $\Omega$; it is a subset of it, as all of its elements are subsets of $\Omega$. So perhaps what you meant to say was that $F \subseteq P(\Omega)$. It is possible that the set of events contains all possible events, but in general this is not the case—it depends on what events we are actually considering. In this case, $F$ would probably contain the set of all $n$-length palindromic strings.

  • $\begingroup$ Thanks. When you say that "in this case $F$ would probably contain the set of all n-length palindromic strings" Does this imply that $F$ has only one event?(the set of all n-length palindromic strings) $\endgroup$ – roy212 May 6 at 21:41
  • $\begingroup$ Well, it’s kind of a strange example I think. You are interested in the probability that you get a palindromic number, so the event you are considering is the event where any of the outcomes that corresponds to the number being palindromic take place, which is just the set of all outcomes that are palindromic strings. So $F$ has only one event, as you say. $\endgroup$ – 雨が好きな人 May 6 at 21:46
  • $\begingroup$ Could you write to what $m$ is equal to? "This is equal". And would it be right to write it as $\Omega = \{d_1d_2d_3...d_n | d_i \in [0,9]\}$ $\endgroup$ – roy212 May 6 at 21:58
  • $\begingroup$ Sorry—I’ve edited it now. Yes, that seems to be quite an elegant way of describing the set. $\endgroup$ – 雨が好きな人 May 6 at 22:02

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