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.

3 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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.

- 3 years ago

Oh!Thank you sir!I am relieved!sigh

- 3 years ago

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

- 3 years ago

How did you get this answer sir?

- 3 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=$$\dfrac{20 \times 540}{6^6}=\dfrac{25}{108}$$

- 3 years ago

Same thing sir!Thank u!

- 3 years ago