Question

A box contains $6$ red marbles numbers from $1$ through $6$ and $4$ white marbles $12$ through $15$. Find the probability that a marble drawn 'at random' is white and odd numbered.

• A. $5$
• B. $\frac {1}{5}$
• C. $6$
• D. $\frac {1} {6}$

Answer 1

Answer: (B)

$\frac {1}{5}$

Answer 2

Total number of marbles, $n(S)=6+4=10$
Marbles which are white and odd numbered $=2$
$\therefore n(E)=2$
Now, $P(E)=\frac{n(E)}{n(S)}=\frac{2}{10}$
$=\frac{1}{5}$