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A department store is holding a drawing to give free shopping sprees to two lucky customers. There are $18$ customers who have entered the drawing: $4$ live in the town of Gaston, $8$ live in Pike, and $6$ live in Wells. In the drawing, the first customer will be selected at random, and then the second customer will be selected at random from the remaining customers. What is the probability that both customers selected are Pike residents?

Report your answer as an exact fraction.

My answer The probability that the first customer lives in Pike is

P1=8/18=4/9

The probability that the second customer lives in Pike is

P2=7/17

The probability that both customers selected are Pike residents is

P1*P2=(4/9)*(7/17)=28/153

Am I Correct?

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    $\begingroup$ Your phrasing is a bit off. "The probability that the second customer lives in Pike" is actually also $\frac{8}{18}$. What you mean to write and what is relevant to the probability calculations in this problem is instead "The probability that the second customer lives in Pike given that the first customer also lives in Pike" which is $\frac{7}{17}$ as you intended. This latter part of the phrase is highly important and should not be left out as doing so completely changes the meaning of the phrase. $\endgroup$ – JMoravitz Apr 10 '18 at 7:40
  • $\begingroup$ You could also say "the probability then that..." $\endgroup$ – Remy Apr 10 '18 at 7:43
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Yes, your reasoning is correct.

Alternatively we want the probability of choosing $2$ of $8$ Pike residents when choosing $2$ of $18$ total residents.

We have

$$\frac{8 \choose 2}{18 \choose 2}\approx0.183$$

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This is a way to formalize your answer.

For $i=1,2$ let $E_i$ be the event that the $i$-th customer lives in Pike.

Then:$$P(E_1\cap E_2)=P(E_1)P(E_2\mid E_1)=\frac8{18}\frac7{17}$$

As you were told in the comments on your question we have $P(E_1)=P(E_2)=\frac8{18}$ but $P(E_2\mid E_1)=\frac7{17}$.

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