A fair dice is rolled 6 times and the outcomes are listed.The probability that the list contains exactly 3 numbers is?\[\]I am getting 25/108 but the answer given in the book is 5/108.

Select the three no. first i.e \(\binom{6}{3}=20\) . then getting exactly three no. in the list is equivalent to Permutation of 6 distinct objects into 3 distinct boxes so that no box remains empty i.e. \(3^6-\binom{3}{1} \times 2^6 +\binom{3}{2} \times 1^6=540\).

So required probability=\(\dfrac{20 \times 540}{6^6}=\dfrac{25}{108}\)

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## Comments

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TopNewestI'm getting the same answer as you. 5/108 just seems too low, since it is most likely that you will end up with 3 numbers.

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Oh!Thank you sir!I am relieved!

sighLog in to reply

The correct answer according to me too is \(\frac{25}{108}\). So the answer given in the book is provided wrong.

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How did you get this answer sir?

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Select the three no. first i.e \(\binom{6}{3}=20\) . then getting exactly three no. in the list is equivalent to Permutation of 6 distinct objects into 3 distinct boxes so that no box remains empty i.e. \(3^6-\binom{3}{1} \times 2^6 +\binom{3}{2} \times 1^6=540\).

So required probability=\(\dfrac{20 \times 540}{6^6}=\dfrac{25}{108}\)

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