Question

An urn contains 2 white, 3 black and 5 red balls. From it 6 balls are drawn with replacement, one at a time.
Let $E_1$ : first one white ball, then 2 black balls and finally 3 red balls are drawn.
$E _2$ : 1 white, 2 black and 3 red balls are drawn so that identical colour balls appear successively, but the succession of colours may be arbitrary.
and $ E_3: 1$ white, 2 black and 3 red balls are drawn in any succession.
If $\frac{P\left(E_1\right)}{x}=\frac{P\left(E_2\right)}{y}=\frac{P\left(E_3\right)}{z}=\frac{9}{4000}$, then find the value of $(x+y+z)$.

Answer 1

$P ( W )=\frac{2}{10} ; P ( B )=\frac{3}{10} ; P ( R )=\frac{5}{10}$
(a) $ P \left( E _1\right)=\frac{2}{10} \cdot \frac{3}{10} \cdot \frac{3}{10} \cdot \frac{5}{10} \cdot \frac{5}{10} \cdot \frac{5}{10}=\frac{9}{4000}$
(b) $ P \left( E _2\right)=3 ! \cdot \frac{9}{4000}=\frac{9 \cdot 6}{4000}=\frac{54}{4000}=6\left(\frac{9}{4000}\right)$
(c)$P\left(E_3\right)=\frac{6 !}{2 ! 3 !} \cdot \frac{9}{4000}=(60) \frac{9}{4000}=\frac{540}{4000}=60\left(\frac{9}{4000}\right) $
$\therefore \frac{P\left(E_1\right)}{1}=\frac{P\left(E_2\right)}{6}=\frac{P\left(E_3\right)}{60}=\frac{9}{4000} $
$\Rightarrow x=1, y=6 \text { and } z=60$
$\text { Hence } x+y+z=67 $