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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.

1 year, 4 months ago

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I'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. · 1 year, 4 months ago

Oh!Thank you sir!I am relieved!sigh · 1 year, 4 months ago

The correct answer according to me too is $$\frac{25}{108}$$. So the answer given in the book is provided wrong. · 1 year, 4 months ago

How did you get this answer sir? · 1 year, 4 months ago

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=$$\dfrac{20 \times 540}{6^6}=\dfrac{25}{108}$$ · 1 year, 4 months ago