box contains 6 red marbles numbered 1 through 6 and 4 white marbl…

Question

box contains $6$ red marbles numbered $1$ through $6$ and $4$ white marble numbered form $12$ through $15$. find the probability that a marble drawn is:

White
White and odd numbered
Even numbered
Red or even numbered.

✓

Answer

We have 6 red marble numbered 1 - 6 and we have 4 white marble numbered 12 - 15 one marble is tobe drawn$\therefore\text{n(s)}=\ ^{10}\text{C}_1$

E be event of getting white marble,

$\therefore\text{n(E)}=\ ^{4}\text{C}_1$
$\therefore\text{P(E)}=\frac{\ ^{4}\text{C}_1}{\ ^{10}\text{C}_1}=\frac{4}{10}=\frac{1}{5}$

E be the event of getting white marble with odd numbered marble.

$\therefore\text{E}=\big\{13,\ 15\big\}$
$\Rightarrow\text{n(E)}=2$
$\text{P(E)}=\frac{2}{10}=\frac{1}{5}$

E be the event of getting even numbered marble.

$\therefore\text{E}=\big\{2,\ 4,\ 6,\ 12,\ 24\big\}$
$\Rightarrow\text{n(E)}=5$
$\text{P(E)}=\frac{5}{10}=\frac{1}{2}$

$E_1$ be the event of getting red marble.

$\therefore\text{P}\text{(E)}_1=\frac{5}{10}$ [as in (ii)]
$\therefore\text{(E}_1\cap\text{E}_2)=\text{even numbered marble}=\big\{2,\ 4,\ 6\big\}$
$\Rightarrow\text{n}(\text{E}_1\cap\text{E}_2)=3$
$\Rightarrow\text{P}(\text{E}_1\cap\text{E}_2)=\frac{3}{10}$
$\therefore$ by law of addition,
$\Rightarrow\text{P}(\text{E}_1\cup\text{E}_2)=\text{P(E}_1)+\text{P(E}_2)-\text{P}(\text{E}_1\cap\text{E}_2)$
$=\frac{6}{10}+\frac{5}{10}-\frac{3}{10}=\frac{8}{10}$
$=\frac{4}{5}$