1. probability

"Assume that the box contains 7 balls: 4 red, 2 green, and 1 yellow. Balls are drawn in succession without replacement, and their colors are noted until both a red and a green ball have been drawn."

How many outcomes are there in the sample space?"

the answer is 25, but i only know how to arrive at this answer by sketching a convoluted diagram or by racking my brain to figure out all the sets (RG, RRG, RRRG etc) and then counting the outcomes.

is there a faster way to figure out the number of outcomes using algebra or some other method? i don't think i'll have enough time on the test to draw a diagram.

thanks

2. Originally Posted by sdsdsd
"Assume that the box contains 7 balls: 4 red, 2 green, and 1 yellow. Balls are drawn in succession without replacement, and their colors are noted until both a red and a green ball have been drawn."

How many outcomes are there in the sample space?"

the answer is 25, but i only know how to arrive at this answer by sketching a convoluted diagram or by racking my brain to figure out all the sets (RG, RRG, RRRG etc) and then counting the outcomes.

is there a faster way to figure out the number of outcomes using algebra or some other method? i don't think i'll have enough time on the test to draw a diagram.

thanks
I don't see too many short cuts here.

3. Hello, sdsdsd!

I agree with Mr. F . . . I see no "formula" approach.
I think Brute Force listing is the only way . . .

Two draws

$\displaystyle \begin{array}{c}RG \\ GR\end{array}$ . . . 2 ways.

Three draws

$\displaystyle \begin{array}{c}RRG \\ RY\!G \\ Y\!RG \\ GGR \\ GY\!R \\ Y\!GR\end{array}$ . . . 6 ways

Four draws

$\displaystyle \begin{array}{c}RRRG \\ RRY\!G \\ RY\!RG \\ Y\!RRG \\ GGY\!R \\ GY\!GR \\ Y\!GGR\end{array}$ . . . 7 ways

Five draws

$\displaystyle \begin{array}{c}RRRRG \\ RRRY\!G \\ RRY\!RG \\ RY\!RRG \\ Y\!RRRG \end{array}$ . . . 5 ways

Six draws

$\displaystyle \begin{array}{c}RRRRY\!G \\ RRRY\!RG \\ RRY\!RRG \\ RY\!RRRG \\ Y\!RRRRG \end{array}$ . . . 5 ways

Total: 25 ways