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I'm trying to do the following probability question involving, I think, the ''amended'' multiplication rule:

A Jar contains 3 red and 5 black balls. What is the probability of drawing 2 red balls simultaneously ?

I used the formula - $P(A \cap B) = P(A)\cdot P(B\mid A)$

I.E. P(Red ball and then a Red ball) = P(Red)$\cdot$P(Red given the first was Red)

$=\frac 3 8 \cdot \frac 2 7$ ?

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$\color{green}{\checkmark}$ Yes. You have it. Your equation is appropriate, $P(R_1\cap R_2) = P(R_1)P(R_2\mid R_1)$, and you measured the terms correctly.

Another way would be to use combinations: ways to select 2 of 3 red balls out of the ways to select any 2 of all 8 balls. Which affirms your conditional probability solution. $$\frac{\binom{3}{2}}{\binom{8}{2}} = \frac{3\cdot 2}{8\cdot 7}= \frac 3 {28}$$

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Yep. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

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Correct.

One way to confirm this works is to look at all the possible outcomes:

There's $8 \times 7 \over 2$ combination of balls to pick.

There's $3 \times 2 \over 2$ combination of red balls you can pick.

${3 \times 2 \over 2} \over {8 \times 7 \over 2}$ is ${3 \over 8} \times {2 \over 7}$

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