# A box contains 6 white balls and 4 red balls. 2 balls are randomly drawn from the box. Find the probability that the second ball selected is red.

A box contains $6$ white balls and $4$ red balls. We randomly (and without replacement) draw $2$ balls from the box. What is the probability that the second ball selected is red?

I know this is a Multiplication Rule problem.

Why doesn't this problem fall under the category of Total Probability Rule?

• One could argue you don't even need the multiplication rule: the probability is simply $\frac{4}{10}.$ – jvdhooft Jun 19 '18 at 8:42
• The slow way of calculating it is to say $F$ is the event the first ball is red and $S$ is the even the second ball is red and then calculate $P(S)=P(F \cap S)+P(F^c \cap S) = P(F)P(S \mid F)+ P(F^c)P(S \mid F^c)$ which might be said to be a total probability rule – Henry Jun 19 '18 at 8:53

Observe that the probability for the second ball to be red equals the probability for the first ball to be red (so $=\frac4{10}$).
• Observe that the probability for the second ball to be red equals the probability for the first ball to be red --- why is that? What if a red was drawn on the 1st occasion? – user366312 Jun 19 '18 at 10:52
• If a red was drawn on first occasion then this decreases the chance for the second to be red: $P(R_2\mid R_1)<P(R_1)$ but on the other hand if the a not a red was drawn on first occasion then this increases the chance for the second to be red: $P(R_2\mid R_1)>P(R_1)$. This together leads to neutrality. Think of it like this: If you draw all $10$ balls one by one and place them in a row. Can you find any reason why e.g. the ball place on spot $3$ should have a smaller or larger probability to be red then the ball on, let's say, spot $7$? Same logic for spot $1$ and spot $2$ (first and second) – drhab Jun 19 '18 at 11:07
• There is a typo in my former comment. It should be $P(R_2\mid R_1^{\complement})>P(R_1)$ (not $P(R_2\mid R_1)>P(R_1)$). – drhab Jun 19 '18 at 11:36