# Given a bag of 3 red marbles, 5 black, and 8 green, what is the probability that?

A bag has:

$3$ red

$5$ black

$8$ green

marbles.

Total of 16 marbles.

You select a marble, and then another one right after. (without replacement).

What is the probability that $both$ are red?

Probability that first pick is red: $\frac{3}{16}$

Probability that second pick is red: $\frac{2}{15}$ (since one ball is removed)

Probability of both marbles being red is: $\frac{3}{16} \cdot \frac{2}{14} = \frac{1}{40}$

How do I do this using combinations only?

• How exactly did that $\frac{2}{15}$ become $\frac{2}{14}$??? Dec 11 '16 at 7:54
• @barakmanos, as $\frac 3{16}\cdot\frac 2{15}=\frac 1{40}$ it looks like a sinple typo. Dec 11 '16 at 7:59

Hint: $$\frac {\text { number of ways in which you can choose 2 red balls without replacement from 15 balls}}{ \text {number of ways in which you can choose 2 balls of any colour without replacement from 15 balls}}=\frac {\binom {3}{2}}{ \binom {16}{2}}=??$$

Hope this helps you.

• thanks alot, it works out! just one question tho, how did u know that sample space size is 16C2, ? in general even, like i understand the top half of fraction, but where did u get bottom half from Dec 11 '16 at 5:51
• It is the general sample space which in this case is the total number of ways in which we can choose 2 balls of any colour from the given 16 balls. Now you must realise that the total sample space is always without any constraint and you need to practice more problems of this kind so that you actually get a feel which tells you how you need to deal with this kind of problems. Dec 11 '16 at 6:03
• @KracX The probability of selecting 2 from the 3 red balls when selecting any 2 from all (3+5+8) balls is: $\frac{^3\mathrm C_2}{^{16}\mathrm C_2}$ Dec 11 '16 at 8:02

Its very simple.

Probability = $\frac{C(3,2)}{C(16,2)}$

$= \frac{\frac{3!}{2!*1!}}{\frac{16!}{14!*2!}}$

$= \frac{3}{120} = \frac{1}{40}$