Question

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

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A $\frac{1}{5}$

Required probability $=$ probability of drawing 1 White or 2 or 3 or 4 or 5 or 6 White $=(\frac{1}{6}\ast \frac{{}_1^{6}C}{{}_1^{10}C})+(\frac{1}{6}\ast \frac{{}_2^{6}C}{{}_2^{10}C})+(\frac{1}{6}\ast \frac{{}_3^{6}C}{{}_3^{10}C})+(\frac{1}{6}\ast \frac{{}_4^{6}C}{{}_4^{10}C})+(\frac{1}{6}\ast \frac{{}_5^{6}C}{{}_5^{10}C})=\frac{251}{1260}$
This is closest to $\frac{1}{5}$