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About Paul Klarreich
Expertise
I can answer questions in basic to advanced algebra (theory of equations, complex numbers), precalculus (functions, graphs, exponential, logarithmic, and trigonometric functions and identities), basic probability, and finite mathematics, including mathematical induction. I can also try (but not guarantee) to answer questions on Abstract Algebra -- groups, rings, etc. and Analysis -- sequences, limits, continuity. I won't understand specialized engineering or business jargon.

Experience
I taught at a two-year college for 25 years, including all subjects from algebra to third-semester calculus.

 
   

You are here:  Experts > Science > Mathematics > Advanced Math > Binomial expansion.

Topic: Advanced Math



Expert: Paul Klarreich
Date: 5/14/2008
Subject: Binomial expansion.

Question
QUESTION: In a pile of 40 oranges, 10 are rotten. What is the probability of getting at least 3 rotten oranges if 20 oranges are picked at random? Please explain..

ANSWER: Questioner:   jagdeep
Category:  Advanced Math
Private:  No
 
Subject:  probability
Question:  In a pile of 40 oranges, 10 are rotten. What is the probability of getting at least 3 rotten oranges if 20 oranges are picked at random? Please explain..
...................................
Hi, jagdeep,

If you pick 20 out of 40, there are C(40,20) possible subsets.  

0. What is p(0)? i.e p(zero rottens)

Of these, C(30,20) will contain no rotten ones.  So

p(0) = C(30,20)/C(40,20)

1. What is p(1)?  You want a subset with 1 rotten, 19 good.

There are C(10,1) 1-element subsets of rottens.
There are C(30,19) 19-element subsets of goods.
The product is C(10,1)C(30,19), so

p(1) = C(10,1)C(30,19)/C(40,20)


2. What is p(2)?  You want a subset with 2 rotten, 18 good.

There are C(10,2) 2-element subsets of rottens.
There are C(30,18) 18-element subsets of goods.
The product is C(10,2)C(30,18), so

p(2) = C(10,2)C(30,18)/C(40,20)

Now the sum of those is p(at most 2), and

p(at least 3) = 1 - p(at most 2)

I think you can work it from there.


---------- FOLLOW-UP ----------

QUESTION: Hi!
Thanks for your help!!
I am still having some doubt with it.
Actually, If I change the question like:
In a pile of oranges, 10% of the oranges are rotten.
What is the probability of getting 3 rotten oranges if 20 ar epicked at random. Also, can you please show me the mathematical calculations also, as I am still getting wrong answer...
Thanks

Answer
Questioner:   jagdeep
Category:  Advanced Math
Private:  No
 
Subject:  probability
Question:  QUESTION: In a pile of 40 oranges, 10 are rotten. What is the probability of getting at least 3 rotten oranges if 20 oranges are picked at random? Please explain..

ANSWER: Questioner:   jagdeep
Category:  Advanced Math
Private:  No

Subject:  probability
Question:  In an INFINITE pile of oranges, 1/40 are rotten. What is the probability of getting at least 3 rotten oranges if 20 oranges are picked at random? Please explain..
...................................
Hi, jagdeep,

---------- FOLLOW-UP ----------

QUESTION: Hi!
Thanks for your help!!
I am still having some doubt with it.
Actually, If I change the question like:
In a pile of oranges, 10% of the oranges are rotten.

>>>> Not 1/4 any more?


What is the probability of getting 3 rotten oranges if 20 ar epicked at random. Also, can you please show me the mathematical calculations also, as I am still getting wrong answer...
Thanks
................................
Yes, this is a different question.  Now, the probability is obtained by the binomial expansion:

In general, if  p = p(a given orange is rotten), and
(1 - p) = p(a given orange is good), then if n oranges are picked, we have:

p(exactly k bad ones) =  C(n,k) p^k (1- p)^(n-k)

naturally,

SUM[k=0 to n] C(n,k) p^k (1- p)^(n-k) = (p + (1-p))^n = 1

0. What is p(0)? i.e p(zero rottens)

p(0) = C(20,0) (9/10)^20

1. What is p(1)?  

p(1) = C(20,1) (9/10)^19 (1/10)^1


2. What is p(2)?

p(2) = C(20,2) (9/10)^18 (1/10)^2

Now the sum of those is p(at most 2), and

p(at most 2) = [9^20 + 20(9^19) + 190(9^18)]/10^20,

which I will leave to you, and

p(at least 3) = 1 - p(at most 2)


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