﻿ Probability of Correct Answers on Multiple Choice Questions

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### 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/
```

### 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.

```

```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

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/
```

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 Ne be the number of collisions per million miles that an Expert Drivers is expected have a single driver auto collision, and Nn 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, Re, is the sum of his likelihood per million miles of having a single driver collision Ne, the square of this probability to represent a two driver collision involving another Expert Driver Ne2, and Ne  times Nn  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:

(Ne +  .7Ne2  +.3 NeNn) x 1,056 billion miles driven = 24,087 collisions

Re = Ne +  .7Ne2  + .3NeNn = 22.8

The equation for Novice Drivers is similar:

(Nn +  .3Nn2  + .7NeNn) x 455 billion miles driven = 17,729 collisions

Rn = Nn +  Nn2  + NeNn = 39

Modified Tuesday, November 17, 2009

Copyright @ 2007 by Fathers' Manifesto & Christian Party