Question

A box contains 6 red marbles numbered 1 through 6 and 4 white marbles numbered from 12 through IS. Find the probability that a marble drawn is

(i) white

(ii) white and odd numbered

(ill) even numbered

(iv) red or even numbered.

Answer 1

Total number of marbles (6 + 4) = 10
Let S be the sample space.

Then n(S) = number of ways of selecting one marble out of $10{=}^{10}{C}_1=10$ ways

(i) Let $E_1$ = event of getting a white marble
$\therefore n\left(E_1\right){=}^4{C}_1=4$

Hence, required probability $\frac{{}^4{C}_1}{{}^{10}{C}_1}=\frac{4}{10}=\frac{2}{5}$

(ii) Let $E_2$ = event of getting a white marble,
which is odd numbered

i.e., $E_2=\{13,15\}$

$\therefore n\left(E_2\right)=2$

(iii) Let E_3 = even of getting an even numbered marble

i.e., $E_3=\{2,4,6,12,14\}\therefore n\left(E_3\right)=5$

Hence required probability=$\frac{n\left(E_3\right)}{n\left(S\right)}$

$=\frac{5}{10}=\frac{1}{2}$

Let $E_4$ = event of getting a red marble

i.e., $E_4=\{1,2,3,4,5,6\}n\left(E_4\right)=6$

Now, $P\left(E_4\right)=\frac{6}{10}=\frac{3}{5}\dots \left(i\right)$

Let $E_5$ = event of getting en numbered marble

Then $E_5=\{2,4,6,12,14\}$

i.e., $n\left(E_5\right)=5$

Now $P\left(E_5\right)=\frac{5}{10}=\frac{1}{2}$

From (i) and (ii), we get:

$E_4\cap E_5=\{2,4,6\}$

$\Rightarrow n(E_4\cap E_5)=3$

$\Rightarrow P(E_4\cap E_5)=\frac{3}{10}$

By addition theorem, we have:

$P(E_4\cap E_5)=P\left(E_4\right)+P\left(E_5\right)-P(E_4\cap E_5)$

$\Rightarrow P(E_4\cup E_5)=\frac{3}{5}+\frac{1}{2}-\frac{3}{10}=\frac{8}{10}=\frac{4}{5}$

Hence, required probability

$=P(E_4\cup e_5)=\frac{4}{5}$