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.
Best Answer
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, $\mathbb{P}(R_1)=\mathbb{P}(R_2)=\mathbb{P}(B_1)=\mathbb{P}(B_2)=\frac{5}{10}=0.5$. $$\therefore \mathbb{P}\Big{(}(R_1\cap B_2)\cup(R_2\cap B_1)\Big{)}=0.5\times0.5+0.5\times0.5=0.5.$$