Question

a jar contains 5 blue marbles and 3 red marbles. suppose you choose a marble at random and do not replace it. then choose a second marble. find the probability of the following event. both of the selected marbles are red.

Answer 1

$\mathbb P(R_1\cap R_2)=\mathbb P(R_2\mid R_1)\mathbb P(R_1)$

The probability of drawing a red marble on the first attempt is

$\mathbb P(R_1)=\dfrac{\dbinom31}{\dbinom51}=\dfrac35$

After the first marble is drawn, and we know it to be red, we're drawing the next marble from a pool with one less marble than before (i.e. conditioning on the event $R_1$). So

$\mathbb P(R_2\mid R_1)=\dfrac{\dbinom21}{\dbinom71}=\dfrac27$

And so

$\mathbb P(R_1\cap R_2)=\dfrac35\cdot\dfrac27=\dfrac6{35}$