Related questions with answers

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This exercise describes an example of a nonmeasurable set in R. If x and y are real numbers in the interval [0, 11, we say that x and y are equivalent, written $x \sim y$ x - y is rational. The relation $\sim$ is an equivalence relation, and the interval [0, 11] can be expressed as a disjoint union of subsets (called equivalence classes) in each of which no two distinct points are equivalent. Choose a point from each equivalence class and let E be the set of points so chosen. We assume that E is measurable and obtain a contradiction. Let $A=\left\{r_{1}, r_{2}, \dots\right\}$ denote the set of rational numbers in [-1, 11 and let $A=\left\{r_{1}, r_{2}, \dots\right\}$ . a) Prove that each $E_n$ is measurable and that $\mu\left(E_{n}\right)=\mu(E)$ . b) Prove that $\left\{{E}_{1}, {E}_{2}, \ldots\right\}$ s a disjoint collection of sets whose union contains [0, 1] and is contained in [-1, 2]. c) Use parts (a) and (b) along with the countable additivity of Lebesgue measure to obtain a contradiction.

Question

Refer to a small consumer survey in which 10 people were asked to choose a cola among Coke, Pepsi, and RC. If each person chose a cola randomly, what is the probability that at least one person did not choose Coke?

Solution

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Step 1

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$\textbf{Multiplication principle: }$ If one event can occur in $m$ ways AND a second event can occur in $n$ ways, then the number of ways that the two events can occur in sequence is then $m\cdot n$ .

There are 3 options for each of the 10 people (Coke, Pepsi, and RC) and thus there are $3^{10}$ possible outcomes by the multiplication principle.

\text{\# of possible outcomes}=3^{10}=59049

The order in which we select the Coke answers does not matter (as a different order results in the same people answering Coke) and thus we need to use the definition of $\textbf{combination}$ .

There are $C(10,0)$ ways to select 0 of the 10 people (who answer Coke), thus there are $59049-C(10,0)$ ways to select at least one person how answer Coke and thus there are $59049-C(10,0)$ favorable outcomes.