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Binomial Probability


Date: 04/14/99 at 13:38:21
From: Amy Pacyon
Subject: Probability

I am unfamiliar with the binomial probability formula; it was not 
covered in class. The question is: 

A 10-question multiple choice exam is given, and each question has 
five possible answers. Pasxal takes this exam and guesses at every 
question. Use the binomial probability formula to find the probability 
(to 5 decimal places) that 

A) he gets exactly 2 questions correct; 
B) he gets no questions correct; 
C) he gets at least one question correct (use the information from 
   part B to answer this part); 
D) he gets at least 9 questions correct; 
E) without using the binomial probability formula, determine the 
   probability that he gets exactly 2 questions correct; 
F) Compare your answers to parts a and f.  If they are not the same 
   explain why.

Thank you in advance for your help.


Date: 04/14/99 at 15:03:23
From: Doctor Anthony
Subject: Re: Probability

You say that you have not yet covered binomial probability, so I will 
start with a general note on the topic.

Flipping coins is an example of binomial probability, where there are 
repeated trials, with constant probability of success (say getting a 
head) at each trial. As an example suppose there are 4 with 
probability (1/2) of success at each trial.

In the general binomial model if we carry out n trials, with 
probability p of success at each trial and probability q of failure, 
where p+q = 1, then if we want a probability of say r successes, one 
possible sequence is r successes followed by n-r failures.

The probability of this sequence is  ppppp to r terms x qqqqq to n-r 
terms.

However, we can get r successes in a whole range of sequences, the 
number of sequences being the same as the number of possible sequences 
of r p's and n-r q's.

                               n!
The number of sequences  =  --------      = C(n,r)
                             r! (n-r)! 

So the total probability of r successes and n-r failures is   

   P(r) =  C(n,r) p^r q^(n-r)

Returning to our problem with 4 coins, we have n = 4, p = 1/2, and
q = 1/2, and we require a probability of exactly 2 heads.  So r = 2.

   P(2 heads) = C(4,2) (1/2)^2 (1/2)^2

              =  6 x 1/4 x 1/4

              =   6/16

              =  3/8

Now we look at the question of the multiple choice exam.  I will work 
through part (A) and leave you to try and complete the question.

>A) he gets exactly 2 questions correct. 

In this example n = 10,  p = 1/5,  q = 4/5,  r = 2

  P(2) = C(10,2) x (1/5)^2 x (4/5)^8  =  0.3019899

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations
High School Probability

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