Question

Stats (!!) Intro

Answer 1

meant this:

Answer 2

so you randomly choose anything in : \(000 \to 999\)

Answer 3

duh, my bad :( . shoulda done the qu first.

so, for 3-digits, you choose any number in range: \(\color{red}{100} \to 999\)

and then:

- units must be: \(0 \to 9\), ie **10** choices \(~~ \star\)

- tens must be: \(0 \to 9\), ie 10 choices

- hundreds must be: \(1 \to 9\), ie **9** choices \(~~ \triangle\)



In total, you can randomly choose any one of \(10 \times 10 \times 9 = 900\) numbers

If you sift through all 900 possible numbers, disregarding all but the ones with a unit digit that = 7, you discard \( \dfrac{9}{10}\) of the numbers, and retain \( \dfrac{1}{10}\), on the basis of \(\star\), ie you retain 90 possible numbers

So there are 90 numbers with a unit digit that = 7.

So \(P(\text{units digit = 7}) = \dfrac{90}{900} = 0.1\)

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Addressing the 2nd part, if you sift through all 900 possible numbers, disregarding all but the ones with a hundreds digit that = 7, you discard \( \dfrac{8}{9}\) of the numbers, and retain \( \dfrac{1}{9}\), on the basis of \(\triangle\), ie you retain 100 possible numbers

So \(P(\text{hundreds digit = 7}) = \dfrac{100}{900} = \dfrac{1}{9}\)