Deriving sum of dependent events

Question

I'm trying to derive formula for dependents events.

Let say we have a box with 5 balls: 2 whites and 3 red. Calculate the probability of:

• get white first, then red
• OR get red first, then white

As I understand, we should use formula P(A + B), where:

• event A: get white ball
• event B: get red ball
• event (B|A): get red ball if white already taken
• event (A|B): get white ball if red already taken

So, following events can happen:

$${\above 0pt A * (B|A)}$$ $${\above 0pt B * (A|B)}$$ $${{\above 0pt A \space * }{ \above 1pt {(B|A)}}}$$ $${{\above 0pt B \space * }{ \above 1pt {(A|B)}}}$$

Therefore:

$${P(A + B) = P(A) * P(B|A) + P(B) * P(A|B) + P(A) * (1 - P(B|A)) + P(B) * (1 - P(A|B))}$$ $${ = 2 * P(A) * P(B|A) + P(A) - P(A) * P(B|A) + P(B) - P(B) * P(A|B)}$$ $${ = 2 * P(A) * P(B|A) + P(A) - 2 * P(A) * P(B|A) + P(B)}$$ $${ = P(A) + P(B)}$$

P(A) = 2/5 P(B) = 3/5

P(A + B) = 2/5 + 3/5 = 1

Question: Is it right formula and probability?

Answer 1

Let $W_1$ denote the event that the first ball drawn is white.

Let $W_2$ denote the event that the second ball drawn is white.

Let $R_1$ denote the event that the first ball drawn is red.

Let $R_2$ denote the event that the second ball drawn is red.

Then to be found is: $$\begin{aligned}P\left(\left(W_{1}\cap R_{2}\right)\cup\left(R_{1}\cap W_{2}\right)\right) & =P\left(W_{1}\cap R_{2}\right)+P\left(R_{1}\cap W_{2}\right)\\ & =P\left(R_{2}\mid W_{1}\right)P\left(W_{1}\right)+P\left(W_{2}\mid R_{1}\right)P\left(R_{1}\right)\\ & =\frac{3}{9}\frac{2}{10}+\frac{2}{9}\frac{3}{10}\\ & =\frac{2}{15} \end{aligned} $$

Now check yourself.

Answer 2

When you draw two balls, there are four outcomes: $$P(W_1\cap W_2)=P(W_1)\cdot P(W_2|W_1)=\frac25\cdot \frac14=\frac{1}{10};$$ $$P(W_1\cap R_2)=\frac25\cdot \frac34=\frac{3}{10};$$ $$P(R_1\cap W_2)=\frac35\cdot \frac24=\frac{3}{10};$$ $$P(R_1\cap R_2)=\frac35\cdot \frac24=\frac{3}{10}.$$ And you want to find: $$P((W_1\cap R_2)\cup (R_1\cap W_2))=P(W_1\cap R_2)+P(R_1\cap W_2).$$