On a 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.
On a 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.The die is then rolled.Prove that the probability that the top face has an odd number of dots is $\frac{11}{21}.$
Since one of the 21 dots is removed,so either the face having 6 dots become 5 dotted face or the face having 5 dots become 4 dotted face or the face having 4 dots become 3 dotted face or the face having 3 dots become 2 dotted face or the face having 2 dots become 1 dotted face or the face having 1 dot become 0 dotted face.
Probability that the face having 6 dots become 5 dotted face$=\frac{6}{21}$
Probability that the face having 4 dots become 3 dotted face$=\frac{4}{21}$
Probability that the face having 2 dots become 1 dotted face$=\frac{2}{21}$
Now the probability of top face having an odd number of dots is$\frac{1}{2}\times \left(\frac{6}{21}+\frac{4}{21}+\frac{2}{21}\right)$
But this does not give me the correct answer.Where have i gone wrong?Please help me.
$\endgroup$ 11 Answer
$\begingroup$You went wrong on the line beginning with "Now the probability $\ldots$". Actually I cannot see how you arrived at the formula with the $\times$.
I'd argue as follows: With probability ${6\over21}+{4\over21}+{2\over21}={4\over7}$ you change an "even face" to an "odd face" and therewith create a die having four "odd faces", and with probability ${5\over21}+{3\over21}+{1\over21}={3\over7}$ you change an "odd face" to an "even face" and therewith create a die having only two "odd faces". The probability that, after the die has been thrown, the top face has an odd number of eyes is therefore given by $${4\over7}\cdot{2\over3}+{3\over7}\cdot{1\over3}={11\over21}\ .$$
$\endgroup$ 6
