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Q.

On normal standard die one of the 21 dots from any one of the six faces is removed at random with each dot equally likely to be chosen. If the die is then rolled, then the probability that the odd number appears is

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a

$\frac{11}{21}$

b

$\frac{10}{21}$

c

$\frac{4}{21}$

d

$\frac{22}{43}$

answer is A.

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Detailed Solution

$P (E_{1}) = \frac{9}{21}, P (E_{2}) = \frac{12}{21} P (E) = P (E \cap E_{1}) + P (E \cap E_{2}) = \frac{11}{21}$

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