• October 23rd 2006, 05:58 AM
jai
A bag contains (n+7) tennis balls.
n of the balls are yellow
The other 7 balls are white

John will take at random a ball from the bag.
He will look at its colour and then put it back in the bag.

After john has put the ball back into the bag, Mary will then take at random a ball from the bag.
She will note its colour.

Given that the probability than John and Mary will take balls with different colours is 4/9,
Prove that 2n2 – 35n + 98 = 0
• October 23rd 2006, 06:41 AM
Soroban
Hello, jai!

Quote:

A bag contains $n+7$ tennis balls:
$n$ of the balls are yellow, the other $7$ balls are white

John will take at random a ball from the bag.
He will note its colour and then put it back in the bag.

Then Mary will take at random a ball from the bag.
She will note its colour.

Given the probability than John and Mary will take balls with different colours is $\frac{4}{9}$,
. . prove that: . $2n^2 - 35n + 98 \:= \:0$

$\begin{array}{cc}P(\text{1st Y}) \: = \: \frac{n}{n+7} \\ P(\text{2nd W}) = \frac{7}{n+7} \end{array}\quad\Rightarrow\quad P(\text{Y, then W}) = \frac{7n}{(n+7)^2}$

$\begin{array}{cc}P(\text{1st W}) \: = \: \frac{7}{n+7} \\ P(\text{2nd Y}) = \frac{n}{n+7} \end{array}\quad\Rightarrow\quad P(\text{W, then Y}) = \frac{7n}{(n+7)^2}$

$P(\text{different colours})\:=\:\frac{7n}{(n+7)^2} + \frac{7n}{(n+7)^2} \:=\:\frac{14n}{(n+7)^2}$

We are told that this probability is $\frac{4}{9}$

Hence: . $\frac{14n}{(n+7)^2} \:=\:\frac{4}{9}\quad\Rightarrow\quad 126n\:=\:4(n+7)^2\quad\Rightarrow\quad 4n^2 - 70n + 196 \:= \:0$

Divide by 2: . $\boxed{2n^2-35n + 98 \:=\:0}$