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

Basic Definitions Related to Probability

Question

$\frac{1}{5}$

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{4}{5}$

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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 $\frac{1}{5}$