1. ## Probability problem help?

Question:

There are 2 red balls and 4 yellow ones in a box.
a) if a ball is drawn and replaced,and a second ball is drawn,what is the probability that the first ball is red and the second ball is yellow?
b)What is the probability that the balls will be red and yellow combination in any order?

Attempt:
a) total no. of balls =6

2P1+4P1 =6
$\frac{6}{6} =1$

b) 2C1+4C1=6
$\frac{6}{6} =1$

Are my answers correct because I doubt it ?

thank you

2. Originally Posted by mj.alawami
There are 2 red balls and 4 yellow ones in a box.
a) if a ball is drawn and replaced,and a second ball is drawn,what is the probability that the first ball is red and the second ball is yellow?
b)What is the probability that the balls will be red and yellow combination in any order?
a) $\frac{2}{6} \cdot \frac{4}{6}$

b) $2\left(\frac{2}{6} \cdot \frac{4}{6}\right)$

Now WHY?

3. Originally Posted by Plato
a) $\frac{2}{6} \cdot \frac{4}{6}$

b) $2\left(\frac{2}{6} \cdot \frac{4}{6}\right)$

Now WHY?
Shouldn't the first answer be related to permutation since they are in order and the second is combinations ?

4. Originally Posted by mj.alawami
Shouldn't the first answer be related to permutation since they are in order and the second is combinations ?
It has almost nothing to to do with permutations or combinations.

a) There are two ways to pick a red ball out six. Then there are four ways to pick a yellow ball out six.

b) is just twice a). Because we could have $RY \text{ or }YR$.

Don't overthink this sort of question.