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)=(\dbinom31)/(\dbinom51)=\frac35`

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)=(\dbinom21)/(\dbinom71)=\frac27`

And so


`\mathbb P(R_1\cap R_2)=\frac35\cdot\frac27=\frac6{35}`