1. ## Prove equations.

Given:

the probability a coin lands on heads 3 times out of four tosses = $4p^3(1-p)$

the probability a coin lands on heads four times out of six tosses = $15p^4(1-p)^2$
the probability a coin lands on heads 3 times out of four tosses is twice the probability a coin lands on heads four times out of six tosses. Prove p must satisfy the equation $15p^2-15p+2=0$
thanks.

2. $30p^4(1-p)^2=4p^3(1-p)$

$30p^4(1-p)=4p^3$

$30p^4-30p^5-4p^3=0$

$30p^5-30p^4+4p^3=0$

divde by $p^3$

3. All of these follow from the basic definition of the binomial probability distribution. Do you know what that is?

4. Originally Posted by BabyMilo
Given:

thanks.
first, do not put questions in quotes. notice that they do not show up when i quote you? this makes it annoying.

anyway, the trick here is to realize that you have a binomial distribution here.

Lets say you are repeating an experiment where each outcome is independent of the others. moreover, each outcome can be considered as a "success" or "failure". Then the probability of $k$ successes in $n$ trials, is given by:

$P(k) = {n \choose k} p^kq^{n - k}$

where $p$ is the probability of success and $q = 1 - p$ is the probability of failure, and ${n \choose k} = _nC_k$.

Now, for your first problem: $n = 4, ~k=3$ and for the second, $n = 6,~k = 4$. plug it in the formula i gave you to derive the desired expressions.

for the last problem, set the expression in the first problem equal to 2 times the expression in the second and simplify to get to what you want

5. Originally Posted by Jhevon
according to the problem, the left side of your first equations should be multiplied by 2.
can i just ask why left hand side and not right hand side?

thanks

6. Originally Posted by Jhevon
first, do not put questions in quotes. notice that they do not show up when i quote you? this makes it annoying.

anyway, the trick here is to realize that you have a binomial distribution here.

Lets say you are repeating an experiment where each outcome is independent of the others. moreover, each outcome can be considered as a "success" or "failure". Then the probability of $k$ successes in $n$ trials, is given by:

$P(k) = {n \choose k} p^kq^{n - k}$

where $p$ is the probability of success and $q = 1 - p$ is the probability of failure, and ${n \choose k} = _nC_k$.

Now, for your first problem: $n = 4, ~k=3$ and for the second, $n = 6,~k = 4$. plug it in the formula i gave you to derive the desired expressions.

for the last problem, set the expression in the first problem equal to 2 times the expression in the second and simplify to get to what you want
sry i thought it would be easier to read.

7. Originally Posted by HallsofIvy
All of these follow from the basic definition of the binomial probability distribution. Do you know what that is?
yes this is of a Statistics module. thanks!

8. Yes,

in the question, it's stated,
the probability of the 3 out of 4 case is twice the probabilty of the 4 out of 6 case.

Therefore

$4p^3(1-p)$

is twice the other.

9. Originally Posted by BabyMilo
can i just ask why left hand side and not right hand side?

thanks
did you read the problem? it told us one was twice the other. there is a specific one that should be multiplied by 2.

10. Originally Posted by Archie Meade
Yes,

in the question, it's stated,
the probability of the 3 out of 4 case is twice the probabilty of the 4 out of 6 case.

Therefore

$4p^3(1-p)$

is twice the other.
then shouldnt it be $15p^4(1-p)^2=2(4p^3(1-p))$

11. Originally Posted by BabyMilo
then shouldnt it be $15p^4(1-p)^2=2(4p^3(1-p))$
if you say "A is twice B" you are saying "A = 2B"

now look at what your A and B are.