An urn contains 6 white and 4 black balls. A fair die whose faces are numbered from 1 to 6 is rolled and number of balls equal to that of the number appearing on the die is drawn from the urn at random. The probability that all those are white is

Engineering

Mathematics

Conditional Probability

Question

$\frac{1}{5}$

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{4}{5}$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Required probability = probability of drawing 1 White or 2 or 3 or 4 or 5 or 6 White

$= (\frac{1}{6} \times \frac{_{1}^{6} C}{_{1}^{10} C}) + (\frac{1}{6} \times \frac{_{2}^{6} C}{_{2}^{10} C}) + (\frac{1}{6} \times \frac{_{3}^{6} C}{_{3}^{10} C}) + (\frac{1}{6} \times \frac{_{4}^{6} C}{_{4}^{10} C}) + (\frac{1}{6} \times \frac{_{5}^{6} C}{_{5}^{10} C}) = \frac{251}{1260}$
This is closest to $\frac{1}{5}$

Please subscribe our Youtube channel to unlock this solution.