1. ## probability problem

if any on the can help me with this problem please , thank you

if Obama and Hillary are competing for the democratic nomination,at the democratic convention the party leaders are trying to keep Hilary supporters outside in order to maintain the image of the party unity.if the have done their job all the state delegates will vote for Obama but if they messed up and Hillary supporters get inside 45% of the votes will go to Hillary

when you watch the delegates start to vote in random order.how many delegates do you have to see to know if the party leaders did thier job and keep it all Obama affairs and to be at least 99% certain of your conclusion?

2. Originally Posted by spywx
if any on the can help me with this problem please , thank you

if Obama and Hillary are competing for the democratic nomination,at the democratic convention the party leaders are trying to keep Hilary supporters outside in order to maintain the image of the party unity.if the have done their job all the state delegates will vote for Obama but if they messed up and Hillary supporters get inside 45% of the votes will go to Hillary

when you watch the delegates start to vote in random order.how many delegates do you have to see to know if the party leaders did thier job and keep it all Obama affairs and to be at least 99% certain of your conclusion?
OK, I'll suggest you consider a similar problem.

Suppose you have two bags. Bag 1 has 55 orange balls and 45 hwite balls. Bag 2 has 100 orange balls and zero hwite balls.

You pick a bag at random and remove n balls without replacement. How many balls do you need to remove before being 99% confident that you picked bag 2?

To get you started:

Let m be the number of balls removed.

On the one hand: Pr(Bag 2 | no hwite balls removed in m drawings) $\displaystyle \geq 0.99$.

On the other hand:

Pr(Bag 2 | no hwite balls removed in m drawings)

= Pr( Bag 2 and no hwite balls removed in m drawings)/Pr(no hwite balls removed in m drawings)

= 0.5/Pr(no hwite balls removed in m drawings).

But Pr(no hwite balls removed in m drawings) = Pr(no hwite balls removed in m drawings | bag 1) Pr(Bag 1) + Pr(no hwite balls removed in m drawings | bag 2) Pr(Bag 2) = (......) (0.5) + 0.5.

Therefore $\displaystyle \frac{0.5}{(......) (0.5) + 0.5} = \frac{1}{(.......) + 1} \geq 0.99 \Rightarrow$ ......

PS: Try finding colors that don't have a political or social connotation! If anyone can suggest the name of a color starting with the letter H (that's not too unusual), they'll get a thankyou. Otherwise y'all stuck with hwite.

3. i still cant find the answer. if anyone can help plz .

4. Originally Posted by spywx
i still cant find the answer. if anyone can help plz .
So where are you stuck?

Getting an expression for Pr(no hwite balls removed in m drawings | bag 2)? Or solving the resulting inequality?

5. i do understand what you did but still how all of that will get the number of balls ? ..i know that sound stupid but i am bad with probability problems

6. Originally Posted by mr fantastic
OK, I'll suggest you consider a similar problem.

Suppose you have two bags. Bag 1 has 55 orange balls and 45 hwite balls. Bag 2 has 100 orange balls and zero hwite balls.

You pick a bag at random and remove n balls without replacement. How many balls do you need to remove before being 99% confident that you picked bag 2?

To get you started:

Let m be the number of balls removed.

On the one hand: Pr(Bag 2 | no hwite balls removed in m drawings) $\displaystyle \geq 0.99$.

On the other hand:

Pr(Bag 2 | no hwite balls removed in m drawings)

= Pr( Bag 2 and no hwite balls removed in m drawings)/Pr(no hwite balls removed in m drawings)

= 0.5/Pr(no hwite balls removed in m drawings).

But Pr(no hwite balls removed in m drawings) = Pr(no hwite balls removed in m drawings | bag 1) Pr(Bag 1) + Pr(no hwite balls removed in m drawings | bag 2) Pr(Bag 2) = (......) (0.5) + 0.5.

Therefore $\displaystyle \frac{0.5}{(......) (0.5) + 0.5} = \frac{1}{(.......) + 1} \geq 0.99 \Rightarrow$ ......
[snip]
Originally Posted by spywx
i do understand what you did but still how all of that will get the number of balls ? ..i know that sound stupid but i am bad with probability problems
Pr(no hwite balls removed in m drawings | bag 2) = $\displaystyle \frac{{45 \choose 0} {55 \choose m} }{{100 \choose m} } = \frac{{55 \choose m} }{{100 \choose m} }$.

Therefore you need to find the smallest integer value of m that solves the inequality

$\displaystyle \frac{1}{\frac{{55 \choose m} }{{100 \choose m} } + 1} \geq 0.99$ $\displaystyle \Rightarrow \frac{{100 \choose m}}{{55 \choose m} + {100 \choose m}} \geq 0.99$ .....

I get m = 8. So the minimum nmber of delegates you'd need to see is 8.

Now a question for you. I assumed for simplicity that the bags each had 100 balls. Does the answer for m depend on the number of balls in the bag? In other words, for your problem does it matter whether or not you're told how many state delegates are voting ....?