Q92 / 180 · KCET 2015 · Mathematics

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
    \( \frac{1}{7} \)
  • Correct answer
    B
    \( \frac{1}{5} \)
  • C
    \( \frac{1}{9} \)
  • D
    \( \frac{1}{6} \)

Solution

Number of red marbles are $6(1$ to 6$)$, number of white marbles are $4(12$ to 15$)$. So, R1, R2, R3, R4, R5, R6, W12, W13, W14, W15 Total number of marbles are 10. We have, white + odd = W13, W15 So, required probability is $\frac{2}{10}=\frac{1}{5}$

Hence, the correct answer is option: 'B'.

Relevant Questions