1. ## binomial distribution

please refer to the second line of solution, since we only concerned about the probability of getting number (5) , then why cant I just just say P=(5/6)^5 , why should I times =(5/6)^5 with (1/6)^2 ?

2. ## Re: binomial distribution

Originally Posted by delso
please refer to the second line of solution, since we only concerned about the probability of getting number (5) , then why cant I just just say P=(5/6)^5 , why should I times =(5/6)^5 with (1/6)^2 ?
We have strings of length seven made up it numbers on the faces of a die.
The questions is: "how many of those strings contain exactly two 5's"
There $^7C_2$ places for the two 5's. The probability of two 5's is $\left(\frac{1}{6}\right)^2$
The probability of five non-5's is $\left(\frac{5}{6}\right)^5$.

3. ## Re: binomial distribution

what do u mean by We have strings of length seven made up it numbers on the faces of a die.? do you mean we have length of 7 unit consists of 7 dice which each of it with length 1unit?

4. ## Re: binomial distribution

Originally Posted by delso
what do u mean by We have strings of length seven made up it numbers on the faces of a die.? do you mean we have length of 7 unit consists of 7 dice which each of it with length 1unit?
I just don't think you understand any of this.
The outcome $6~3~5~6~4~4~5$ is one favorable case.
ANY outcome $X~X~5~X~X~X~5$ is also a favorable case if $X\ne 5$
There are $\dbinom{7}{2}$ places for the two 5's.

Hence $\dbinom{7}{2}{\left( {\dfrac{1}{6}} \right)^2}{\left( {\dfrac{5}{6}} \right)^5}$