# probability of getting exactly $1$ red ball and $1$ blue ball

Suppose we have $$2$$ boxes, labelled Box 1 and Box 2. Each box contains $$5$$ blue balls and $$5$$ red balls.

If 2 balls have to be selected, $$1$$ from each box, what is the probability of getting exactly $$1$$ red ball and $$1$$ blue ball.

I am totally unfamiliar with the topic "Probability, Independent & Mutually Exclusive Events"; a good explanation and a hint/solution would be appreciated.

• Show your attempts Commented Oct 1, 2020 at 5:42
• My approach was same as of the the guy have answered down below, I just needed a verification. Commented Oct 1, 2020 at 10:46

Let $$R_1,R_2$$ be the events of getting of 1 red ball from box 1 and box 2 respectively. Let $$B_1,B_2$$ be the events of getting of 1 blue ball from box 1 and box 2 respectively.
Required event occurs if $$R_1$$ and $$B_2$$ occur simultaneously or $$R_2$$ and $$B_1$$ occur together and can therefore, be written as $$(R_1\cap B_2)\cup(R_2\cap B_1).$$ The question is $$\mathbb{P}\Big{(}(R_1\cap B_2)\cup(R_2\cap B_1)\Big{)}=?$$
$$(R_1\cap B_2)$$ and $$(R_2\cap B_1)$$ are mutually exclusive events $$\Rightarrow\mathbb{P}\Big{(}(R_1\cap B_2)\cup(R_2\cap B_1)\Big{)}=\mathbb{P}\Big{(}R_1\cap B_2\Big{)}+\mathbb{P}\Big{(}R_2\cap B_1\Big{)}.$$ $$R_1$$ and $$B_2$$ are independent events. Also, $$R_2$$ and $$B_1$$ are independent events $$\Rightarrow\mathbb{P}\Big{(}R_1\cap B_2\Big{)}=\mathbb{P}(R_1)\cdot\mathbb{P}(B_2)\mathrm{~~~and~~~}\mathbb{P}(R_2\cap B_1)=\mathbb{P}(R_2)\cdot\mathbb{P}(B_1).$$ Now, $$\mathbb{P}(R_1)=\mathbb{P}(R_2)=\mathbb{P}(B_1)=\mathbb{P}(B_2)=\frac{5}{10}=0.5$$. $$\therefore \mathbb{P}\Big{(}(R_1\cap B_2)\cup(R_2\cap B_1)\Big{)}=0.5\times0.5+0.5\times0.5=0.5.$$