Q. A bag contains 3 black, 3 white and 2 red balls. One by one, three balls are drawn without replacement. The probability that the third ball is red, is equal to
A) $\frac{2}{56}$
B) $\frac{3}{56}$
C) $\frac{1}{56}$
D) $\frac{14}{56}$
Explanation
(d)
Three possible cases are
[table] Ist draw | IIst draw | IIIst draw ; Red | Non Red | Red ; Non Red | Red | Red ; Non Red | Non Red | Red [/table]
$\therefore$ Required probability
$= \frac{2}{8} \times \frac{6}{7} \times \frac{1}{6} + \frac{6}{8} \times \frac{2}{7} \times \frac{1}{6} + \frac{6}{8} \times \frac{5}{7} \times \frac{2}{6}$
$= \frac{14}{56}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.