Conditional Probability: Papoulis/Pillai Ex 2-13

Question

Question: A box contains white and black balls. When two balls are drawn without replacement, suppose the probability that both are white is 1/3. (A) Find the smallest number of balls in the box.

Solution from the text:

Let a and b denote the number of white and black balls in the box, and $W_k$ the event

$W_k=$ “a white ball is drawn at the kth draw”

We are given that $ P(W_1 \cap W_2) = 1/3 $.

It follows that $$\begin{align} P(W_1 \cap W_2) &= P(W_2 \cap W_1)\\ &= P(W_2 | W_1)P(W_1)\\&=\frac{a-1}{(a-1)+b}\frac{a}{a+b}\\&=1/3 \;\tag1\end{align}$$

The author goes on to state the following:

Because $b > 0$, $$\frac{a}{a+b} > \frac{a-1}{(a-1)+b}$$

We can rewrite $(1)$ as $$\left(\frac{a-1}{(a-1)+b}\right)^2 < 1/3 < \left(\frac{a}{a+b}\right)^2 \: \tag2$$

Which gives the inequalities $$ (\sqrt{3} + 1)b/2 < a < 1+ (\sqrt{3} +1)b/2 \: \tag3$$

And later goes to show that $a = 2$ for $b=1$.

It is the last two inequalities, $(2)$ and $(3)$, that are giving me trouble, as I can’t figure out how the author got there. I have attempted just looking at the denominators (to no avail), as well as multiplying it out (also to no avail). The only thing that got me close to an answer was focusing on $(2)$ without the two exterior terms squared, and looking solely at the left side of the inequality.

Thanks for your time.

Answer 1

Note that $\dfrac{a-1}{a-1+b}<\dfrac{a}{a+b}$. Multiply this inequality by the first term and use (1) for the first inequality in (2); and by the second term for the second inequality in (2). To obtain (3) from (2), take square-roots and simplify.

Answer 2

Starting from this inequality:

$\frac{a}{a+b} > \frac{a-1}{a+b-1} \tag1$

and given that: $\frac{a}{a+b} \cdot \frac{a-1}{a+b-1}=1/3$ and $a,b > 0.$

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Observation:

Observe that $a$ cannot be equal to one, $a \neq 1$, otherwise the probability will be zero, $P(W_1 \cap W_2) = 0$. Therefore, $a$ has to be greater than one, $a \gt 1$.
For $b$, it has to be greater than or equal to one, $b \geq 1.$

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From $(1)$ we can show that
$(\frac{a}{a+b})^2 = \frac{a}{a+b} \cdot \frac{a}{a+b}> \frac{a}{a+b} \cdot \frac{a-1}{a+b-1} > 1/3$

$(\frac{a-1}{a+b-1})^2 = \frac{a-1}{a+b-1} \cdot \frac{a-1}{a+b-1} < \frac{a}{a+b} \cdot \frac{a-1}{a+b-1} < 1/3$

Thus, we reach to inequality number $(2)$ in your question:

$(\frac{a-1}{a+b-1})^2 < 1/3 < (\frac{a}{a+b})^2. \tag2$

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From $(2)$ we have

$i)$ $1/3 < (\frac{a}{a+b})^2$

$ii)$ $(\frac{a-1}{a+b-1})^2 < 1/3$

$i$:-

$1/3 < (\frac{a}{a+b})^2 \Rightarrow (a+b)^2 < 3a^2 \\ a^2+b^2+2ab<3a^2 \Rightarrow 0<2a^2-b^2-2ab$

$0<2a^2-2ab-b^2$

$0<(a - \frac{b(1+\sqrt{3})}{2})(a - \frac{b(1-\sqrt{3})}{2})$

Therefore, it is either
$a<\frac{b(1-\sqrt{3})}{2}$
or
$a>\frac{b(1+\sqrt{3})}{2}$
but $a>0$, hence

$a>\frac{b(1+\sqrt{3})}{2} \tag3$

$ii$:-

$(\frac{a-1}{a+b-1})^2 < 1/3 \Rightarrow 3(a-1)^2 < (a+b-1)^2$

$3(a^2 - 2a +1) < a^2 + b^2 + 2ab -2(a+b)+1$

$3a^2 - 6a +3 - a^2 - b^2 - 2ab +2(a+b)-1<0$

$2a^2 +a (-4-2b) +2b + 2 - b^2 <0$

$(a-\frac{2+b(1+\sqrt{3})}{2})(a-\frac{2+b(1-\sqrt{3})}{2})<0$

$\frac{2+b(1-\sqrt{3})}{2} < a < \frac{2+b(1+\sqrt{3})}{2}$

But we already know from $(3)$ that

$a>\frac{b(1+\sqrt{3})}{2}$

Therefore

$\frac{b(1+\sqrt{3})}{2} < a < \frac{2+b(1+\sqrt{3})}{2}. \tag5$

$a < \frac{2+b(1+\sqrt{3})}{2} = 1+\frac{b(1+\sqrt{3})}{2},$

Finally, we reached to the result

$\frac{b(1+\sqrt{3})}{2} < a < 1+\frac{b(1+\sqrt{3})}{2}. \tag6$

Answer 3

$$a^2 + b^2 + 2ab \lt 3a^2 \\ 2a^2 - b^2 - 2ab \gt 0 \\ a \gt \frac{2b \ \pm \sqrt{4b^2 + 8b^2}}{4} \\ a \gt \frac{2b \ \pm 2b\sqrt{3}}{4} \\ \bbox[yellow]{a \gt \frac{( \sqrt{3} + 1)b}2} \\ 3(a-1)^2 \lt (a-1)^2 + b^2 + 2b(a-1) \\ 2(a-1)^2 - b^2 - 2b(a-1) \lt 0 \\ a-1 \lt \frac{2b \ \pm \sqrt{4b^2 +8b^2}}{4} \\ \bbox[yellow]{a \lt 1 + \frac{( \sqrt{3} + 1)b}2}$$