Expected count of eights in an 11-digit number

Question

An integer with exactly $11$ digits is selected uniformly at random. On average, how many of its digits are $8$ ?

Short answer

$10/9$ . Use indicators and linearity. The leading digit is uniform on $1$ – $9$ , so it is an $8$ with probability $\tfrac19$ ; each of the other ten digits is uniform on $0$ – $9$ , contributing $\tfrac1{10}$ each. Total $1+\tfrac19=\tfrac{10}{9}$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Expectation, Mean, and Variance
Conditional Expectation and Variance Revisited
Sample Spaces, Events, and the Probability Axioms

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