Q:

probability of pulling different color balls from a set

if a box contains six red and four blue balls, then calculate the probability of drawing first a red ball and then, a blue ball.

correct answer= 0.005

A:

$P(1^{st} = red) = \frac {6}{10}$, after that $P(2^{nd} = blue) = \frac 49$.
Thus, total probability $p = \frac 35 * \frac 49 = \frac {4}{15}$
When you have $10$ balls, then the probability of pulling some definite ball is $\frac {1}{10}$. Thus, it's $\frac{1}{10}$ for each of red balls. As events of pulling one or another red ball are independent, then probability of pulling any red ball is the sum of the probabilities of pulling each of them and equals $\frac {6}{10} = \frac {3}{5}$. Reasoning is absolutely the same for the second action - we have $9$ balls, $4$ of them are blue, so the probabilty is $\frac {4}{9}$. Now we have two events $A = $ "the first ball is red", $B = $ "the second ball is blue". We have a formula $P(A \cap B) = P(B|A)P(A)$. Here $P(A \cap B)$ is what you need to find - the probability of pulling red first and then blue. $P(B|A) - "the second ball is blue under condition that the first ball was red", $P(A) - "the first ball is red". Thus you have to multiply the probabilities we calculated above to get an answer.

