Algebra identities

[Sashoye]

Hi
I am looking at a solution to an math problem but I can't get my head around the last step before the solution.

4p+q-(2p-q)^2
= 4p-4p^2 +q-q^2 +4pq
=4p(1-p)+q(1-q) + 4pq ?how did they get from here to the final answer below.
= 4pq+pq+4pq

thanks in advance for any help.

 

[stapel]

4p+q-(2p-q)^2
= 4p-4p^2 +q-q^2 +4pq
=4p(1-p)+q(1-q) + 4pq how did they get from here to the final answer below.
= 4pq+pq+4pq

Since the "4pq" at the end doesn't change, the equality simplifies as:

. . . . .\(\displaystyle 4p(1\, -\, p)\, +\, q(1\, -\, q)\, =\, 4pq\, +\, pq\, =\, 5pq\)

However, the left-hand side of the above equality actually expands as:

. . . . .\(\displaystyle 4p(1\, -\, p)\, +\, q(1\, -\, q)\, =\, 4p\, -\, 4p^2\, +\, q\, -\, q^2\)

...which is not at all the same as what they've provided (namely, "5pq"). So either there's a major "typo" going on here, or there is more to this exercise than just the first line and instructions (such as, maybe, "simplify").

 

[Ishuda]

Hi
I am looking at a solution to an math problem but I can't get my head around the last step before the solution.

4p+q-(2p-q)^2
= 4p-4p^2 +q-q^2 +4pq
=4p(1-p)+q(1-q) + 4pq ?how did they get from here to the final answer below.
= 4pq+pq+4pq

thanks in advance for any help.

Looks like something from sadistics [no, that was on purpose] where you have a binary distribution with probabilities p and q which sum to 1. That is
q = 1-p
and
p = 1-q
So
4p(1-p) = 4pq
and
q(1-q) = qp = pq

 

[Sashoye]

Looks like something from sadistics [no, that was on purpose] where you have a binary distribution with probabilities p and q which sum to 1. That is
q = 1-p
and
p = 1-q
So
4p(1-p) = 4pq
and
q(1-q) = qp = pq

THANK YOU. It is statistics. I am doing a module called applications of probability. Thanks so much for your help.

 

[Sashoye]

Since the "4pq" at the end doesn't change, the equality simplifies as:

. . . . .\(\displaystyle 4p(1\, -\, p)\, +\, q(1\, -\, q)\, =\, 4pq\, +\, pq\, =\, 5pq\)

However, the left-hand side of the above equality actually expands as:

. . . . .\(\displaystyle 4p(1\, -\, p)\, +\, q(1\, -\, q)\, =\, 4p\, -\, 4p^2\, +\, q\, -\, q^2\)

...which is not at all the same as what they've provided (namely, "5pq"). So either there's a major "typo" going on here, or there is more to this exercise than just the first line and instructions (such as, maybe, "simplify").

I was looking at it the same way but I was looking at it incorrectly. The math is from a applications of probability module. so 1-p=q. Thank you for having a look at it or me.

 

[Steven G]

Mind your p's and q's :cool:

Now that was funny!