Why is $\frac{p(1-p)}{1−(1−p)^2}=\frac{1-p}{2-p}$?

Question

I found this solution in an old exam:

$$\frac{p(1-p)}{1−(1−p)^2}=\frac{1-p}{2-p}$$

Without any further explanation. Could someone explain to me how to do this transition?

bottom left : $$1-(1-p)^2 = 1 - 1 + 2p - p^2 = 2p - p^2 = p (2 - p)$$ or $$1-(1-p)^2 = 1^2-(1-p)^2 = (1 + (1-p))(1- (1-p)) = (2-p) p$$ — Michel Billaud, Oct 03 '16 at 10:06

score 14 · Answer 1 · answered Oct 03 '16 at 08:09

14

\begin{align} \frac{p(1-p)}{1−(1−p)^2}&=\frac{p(1-p)}{(1+1-p)(1-1+p)}\\ &=\frac{p(1-p)}{p(2-p)}\\ &=\frac{1-p}{2-p}. \end{align}

answered Oct 03 '16 at 08:09

Yuxiao Xie

8,536

Considering a subtlety proposed by @YvesDaoust, this is only valid when read from left to right. – Yuxiao Xie Oct 03 '16 at 08:50

score 6 · Answer 2 · answered Oct 03 '16 at 08:11

You have two basic options to solve this.

First off, you can just write out the denominator to get $$1-(1-p)^2=1-(1-2p+p^2) = -p^2 +2p = p(2-p)$$

and then cancel out the $p$ top and bottom.

Or, you can "cleverly" see that $1-(1-p)^2 = 1^2-(1-p)^2$ and use the equality $$a^2-b^2 = (a-b)(a+b)$$

to get basically the same thing, since

$$1-(1-p)^2 = (1-(1-p))(1+(1-p)) = p(2-p)$$

score 4 · Answer 3 · 2016-10-03T15:46:31.517

4

By inspection (or by $(1-p)^2=1$), $p=0$ and $p=2$ are roots of the denominator. As the coefficient of the highest power is $-1$, it factors as $p(2-p)$. The rest is up to you.

edited Oct 03 '16 at 15:46

answered Oct 03 '16 at 08:37

Note that the fact that the expressions may be undefined at $p=0$ or $p=2$ does not matter, the factorization of the denominator remains valid. – Oct 03 '16 at 08:41
Yes, factorization remains valid – but reducing $\frac pp$ to $1$ changes the domain. – CiaPan Oct 03 '16 at 09:00
1

@CiaPan: that's why I said "The rest is up to you." – Oct 03 '16 at 09:00

score 3 · Answer 4 · answered Oct 03 '16 at 08:09

3

hint: it is $$1-(1-p)^2=(1-(1-p))(1+1-p)$$

answered Oct 03 '16 at 08:09

Dr. Sonnhard Graubner

95,283

Dietrich Burde · Answer 5 · 2016-10-03T08:31:02.133

1

Perhaps one should say that $p=0$ and $p=2$ are excluded, because then not both terms are defined. For example, $$ \frac{p(1-p)}{1−(1−p)^2}=\frac{0}{0}\neq \frac{1}{2}= \frac{1-p}{2-p}. $$ For $p\neq 0,2$ however, both terms are defined and we can cancel, i.e., $$ \frac{p(1-p)}{1−(1−p)^2}=\frac{p(1-p)}{p(2-p)}=\frac{1-p}{2-p}. $$

edited Oct 03 '16 at 08:31

answered Oct 03 '16 at 08:20

Dietrich Burde

130,978

Well, isn't $p=0$ undefined since then the denominator $1-(1-p)^2=0$? – Yuxiao Xie Oct 03 '16 at 08:22
Yes, of course.. – Dietrich Burde Oct 03 '16 at 08:22
1

By convention, the denominator $\not=0$ and so $p=0,2$ do not need to be taken into consideration. – Yuxiao Xie Oct 03 '16 at 08:25
2

@YuxiaoXie: I agree with you, but there is a little subtlety: for $p=0$, the RHS is defined, so the identity cannot be read "from right to left". – Oct 03 '16 at 08:30

Why is $\frac{p(1-p)}{1−(1−p)^2}=\frac{1-p}{2-p}$?

5 Answers5