Please help!

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.

#Combinatorics #Probability

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Brian Charlesworth - 6 years ago

Oh!Thank you sir!I am relieved!sigh

Adarsh Kumar - 6 years ago

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

Sandeep Bhardwaj - 6 years ago

How did you get this answer sir?

Adarsh Kumar - 6 years 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= $\dfrac{20 \times 540}{6^6}=\dfrac{25}{108}$

Sandeep Bhardwaj - 6 years ago

@Sandeep Bhardwaj – Same thing sir!Thank u!

Adarsh Kumar - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$