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See also http://www.stat.wvu.edu/SRS/Modules/Binomial/test.html
http://mathforum.org/library/drmath/view/56561.html
Multiple Choice Test: Binomial Probability
Date: 08/05/97 at 18:55:12
From: Heather
Subject: Multiple choice test
A multiple choice test consists of 9 questions with 5 choices for each
answer. If a student guesses randomly, find the probability of each of
the following events:
1. The student gets 5 correct
2. The student gets at least 5 correct
3. The student gets 2 correct answers
4. The student gets fewer than 3 correct answers.
Date: 08/06/97 at 11:55:26
From: Doctor Anthony
Subject: Re: Multiple choice test
This is binomial probability with n = 9, p = 1/5, q = 4/5
1. P(5) = 9C5 (1/5)^5 (4/5)^4 = 0.016515
2. At least 5 is = P(5) + P(6) + P(7) + P(8) + P(9)
P(6) = 9C6 (1/5)^6 (4/5)^3 = .0027525 and so on.....
3. P(2) = 9C2 (1/5)^2 (4/5)^7 = 0.301989
4. P(< 2) = P(0) + P(1)
P(0) = (4/5)^9 = 0.1342177
P(1) = 9C1 (1/5) (4/5)^8 = 0.301989
So P(<2) = Sum of these = 0.43621
Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/

http://mathforum.org/library/drmath/view/56189.html
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 10question 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 nr failures.
The probability of this sequence is ppppp to r terms x qqqqq to nr
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 nr q's.
n!
The number of sequences =  = C(n,r)
r! (nr)!
So the total probability of r successes and nr failures is
P(r) = C(n,r) p^r q^(nr)
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/

Probability of Collision
Probability of Collision 
Expert Driver 
Novice Driver 
Total 
Ratio 
Billion miles driven 
1,056 
455 
1,511 
.43 
collisions 
24,087 
17,729 
41,816 
.74 
collisions per billion miles 
22.8 
39 
27.7 
1.71 
Percent of miles driven 
70% 
30% 
100% 
.43 
Percent of collisions 
57.6% 
42.4% 
100% 
.74 
The 71% higher probability per mile that Novice
Drivers will have an automobile collision contributes to an increase in the collision rate
for Expert Drivers. To determine exactly how much higher the collision rate for
Expert Drivers is because of Novice Drivers it's necessary to calculate the rate per one
million miles that both Expert Drivers and Novice Drivers are expected to have an
collision. Let N_{e} be the number of collisions per million miles that an
Expert Drivers is expected have a single driver auto collision, and N_{n }that a
Novice Driver will, and then there are two equations and two variables. The total
number of collisions per million miles that an Expert Driver is expected to have an
collision, R_{e}, is the sum of his likelihood per million miles of having a
single driver collision N_{e}, the square of this probability to represent a two
driver collision involving another Expert Driver N_{e}^{2}, and N_{e}
times N_{n} to represent a two driver collision involving a Novice
Driver. For simplicity, collisions involving more than two drivers are omitted, but
they are rare enough that the ratios below won't change significantly and it's unlikely
that the probability of either driver to have a multiple car crash is much different than
the probability of a two driver collision:
(N_{e} + .7N_{e}^{2}
+.3 N_{e}N_{n}) x 1,056 billion miles driven = 24,087 collisions
R_{e} = N_{e} + .7N_{e}^{2}
+ .3N_{e}N_{n }= 22.8
The equation for Novice Drivers is similar:
(N_{n} + .3N_{n}^{2}
+ .7N_{e}N_{n}) x 455 billion miles driven = 17,729 collisions
R_{n} = N_{n} + N_{n}^{2}
+ N_{e}N_{n} = 39

