Question

In 2 bags there are to be put altogether 5 red & 12 white
balls,neither bag being empty. How must the balls be
divided so as to give a person who draws one ball from
either bag the least chance of drawing a red ball ?

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{ In 2 bags there are to be put altogether 5 red & 12 white }\hspace{.33em}\\~\\
& \normalsize \text{ balls,neither bag being empty. How must the balls be }\hspace{.33em}\\~\\
& \normalsize \text{ divided so as to give a person who draws one ball from}\hspace{.33em}\\~\\
& \normalsize \text{ either bag the least chance of drawing a red ball ?}\hspace{.33em}\\~\\
& a.)\ \dfrac{3}{35} \hspace{.33em}\\~\\
& b.)\ \dfrac{5}{32} \hspace{.33em}\\~\\
& c.)\ \dfrac{7}{32} \hspace{.33em}\\~\\
& d.)\ \dfrac{1}{16} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

I'm very sorry, I don't know your answer

Answer 3

answer given is 5/32 ,

Answer 4

Ya

Answer 5

Yes

Answer 6

Try B

Answer 7

B

Answer 8

this is the closest thing i can come up with on the google

Answer 9

We could reason as follows:
There are 5 red and 12 white.
We have to put at least one ball in each bag.
Say we put a red in one, and nothing in the other, then the probability of drawing a red would be:
(1/1+4/16)/=0.75.

Can we do better?

If we exchange and put a white in bag A, and the rest in the other (B), then
probability =(0+5/16)/2 = 0.15625

At this point, moving any white ball to bag A will increase the probability of drawing in B, so is moving any red.
So the distribution A(0,1) and B(5,11) is a local minimum giving P(R)=5/32.

I have not yet found a way to show that this is also the global minimum without brute force (which is what I did). Hope someone would give it a try.

Answer 10

in book it is given ,

\(\large \color{black}{\begin{align}
& \dfrac12 \times \dfrac{5}{16}+ \dfrac12 \times 0= \dfrac{5}{32} \hspace{.33em}\\~\\
\end{align}}\)

which i didnt understand

Answer 11

Yes, that's the same as :
(0+5/16)/2 = 0.15625 = 5/32