2 different ways to solve a probability problem

[r_walid]

I recently posted the following question:

There are 5 red balls and 10 blue balls.
if one is to pick 5 balls, What is the probability of picking 2 red and 3 blue?

The answer I recieved which I am convinced of is the following:
Probability = no. of desired outcomes / total no. of outcomes.

no.of desired outcomes = (5c2) x (10c3)
total no. of outcomes = (10+5)c5 = 15c5

Probability = (5c2 x 10c3)/ 15c5 = 400/1001



My question is this. I tried solving the problem in a different way but i'm getting a different result. can someone explain to me why?

Here is my reasoning.
P(2 red and 3 blue ) = P(2 red) x P(3 blue).

To pull the first 2 red balls, the prob = 5C2/15C2.
To have 3 blue balls in my next pick, the prob would equal= 10C3 / 13C3

it's 13C3 since i've already picked 2 red balls before.

so when you do 5C2/15C2 x 10C3 / 13C3.

After calculation, the number i get is different from the previous answer: 400/1001

What am I doing wrong?

[palvarez]

I recently posted the following question:

There are 5 red balls and 10 blue balls.
if one is to pick 5 balls, What is the probability of picking 2 red and 3 blue?

The answer I recieved which I am convinced of is the following:
Probability = no. of desired outcomes / total no. of outcomes.

no.of desired outcomes = (5c2) x (10c3)
total no. of outcomes = (10+5)c5 = 15c5

Probability = (5c2 x 10c3)/ 15c5 = 400/1001



My question is this. I tried solving the problem in a different way but i'm getting a different result. can someone explain to me why?

Here is my reasoning.
P(2 red and 3 blue ) = P(2 red) x P(3 blue).

To pull the first 2 red balls, the prob = 5C2/15C2.
To have 3 blue balls in my next pick, the prob would equal= 10C3 / 13C3

it's 13C3 since i've already picked 2 red balls before.

so when you do 5C2/15C2 x 10C3 / 13C3.

After calculation, the number i get is different from the previous answer: 400/1001

What am I doing wrong?

In the second case, you forgot to consider various combinations of "2 R 3B".

If you consider them, you will end up with:

(5!/(3!2!)) (5c2/15c2)*(10c3/13c3)

the above is same as your first method.