Probability picking red and white balls

Q1) Answer these questions
What is the probability of picking a red ball on the first draw?
How many red balls are left?
How many total balls are left?
What is the probability of picking a red hall on the second draw?
How many red balls are left?
How many total balls are left?
Repeat until you get to through the 5th red ball draw

Now think of flipping a coin--the probability of getting heads on a two faced coin is 1 out of two. The probability of getting two heads out of two flips is one out of four. Three heads out of three flips is one out of eight, four out of four is one out of sixteen, five out of five 1/2(1/16)=1/32.

Use this principal to predict predict the five consequitive red balls out of the balls pot

HINT you aren't replacing the red balls--if you were, the probability of 5 consequitive red ball draes would be (1/5)^5.

Rap

Thank you. I figured out Q1. (20/100)*(19/99)*(18/98)*(17/97)*(16/96)=0.0002

Now, do you have an idea on Q2? Thanks again.

Think of the number of ways to get 2 combinations of 5 red ball draws out of 500 draws .

You can do this 2 ways--count the possibilities, or develope a method of counting.

Lets do a developement based upon counting possibilities
2 out of 2 draws-- 1st # 2nd , one way

2 out of three draws 1st & 2nd, 1st & 3rd, 2nd & 3rd --3 ways
I'm gonna use a shorthand from now on possibilities this case is 1/2, 1/3. and 2/3

2 out of four draws--- 1/2. 1/3. 1/4. 2/3. 2/4. 3/4 ---6 ways

2 out of 5 draws---1/2. 1/3. 1/4. 1/5. 2/3. 2/4. 2/5. 3/4. 3/5. 4/5--10 ways

I'm beginning to see a pattern

2 out of 6 draws--10+5 ---15 ways

2 out of 7---15+6---21 ways

These are called triangular numbers and they are the sum of the (n-1)th possibilities. The Greeks called these triangular numbers and proved by induction that they can be generally calculated by the formula (n-1)*n/2

that is 2 out of 2=1*2/2=1
2 out of 2=2*3/2=3
2 out of 4=3*4/2=6
2 out of 5=4*5/2=10
2 out of 6=5*6/3=15
2 out of 7=6*7/2=21
....
2 out of n=(n-1)*n/2

let n be 500 and you'll get the number of ways you can get 2 consequitive red balls out of 500

BTW this was further generalized in the 16th Century from combinations of k oblects out of n possibilities which is given by n!/[k!(n-k)!].
In this case k=2 and n= number of draws---this collapses to n(n-1)/2 which is same as the triangular number.

So you now know the ways you can get 2 sets of consequitive red balls--and the probability of pulling one set of 5 consequitive red balls (from the first question). Call this P'(e)

But you're asking it to be done twice--so I gotta take it to that power, that is P'(e)*P'(e)

Now multiply that probability by the number of ways you can get 2 combinations out of n draws and you'll answer this question.

Rap

Part (B)
Let p = the answer from part (A).
P(twice or more) = 1 - P(zero) - P(once)
P(zero) = (1-p)^500
P(once) = 500*p*(1-p)^499

So here's my interpretation based on your awesome explanation:

step 1:
Probability of pulling one set of 5 consequitive red balls is (20/100)*(19/99)*(18/98)*(17/97)*(16/96)=0.0002

step 2:
All possible combination of pulling 2 sets of 5 consequitive red balls from 500 draws is 500(500-1)/2 = 124750

step 3:
Because I want to find the probability of pulling 2 sets of 5 consequitive red balls, I need to first do 0.0002*0.0002 = 0.00000004

step 4:
Final probability would be 0.00000004*124750=0.00499

I'm sorry but I'm a bit confused on step 4 because the final probability seems high from a common sense stand point. The probability to get 5 consequitive red balls is already low (1 out of 5000). Shouldn't the final probability be very low if I'm trying to get 5 consequitive red balls twice within 500 draws? Or am I thinking in the wrong direction?

Thank you again for your time. I really appreciate it.

Rap's method for part two is wrong. Use mine. His method doesn't account for the probabilities associated with the 498 times you don't get 5 red balls. Nor does it account for getting 5 red balls more than twice.

It does come close (0.00529), but you rounded too soon so it appears closer than it really is (0.00494).

Markr is correct--my method does not account for the probability of getting 5 consequitive ball more than twice out of 500 draws.

thank you very much.