SOLUTION: A quiz consists of 20 multiple-choice questions, each with 4 possible answers. For someone who makes random guesses for all of the answers, find the probability of passing if the m

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Question 1206327: A quiz consists of 20 multiple-choice questions, each with 4 possible answers. For someone who makes random guesses for all of the answers, find the probability of passing if the minimum passing grade is 70 %.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
A quiz consists of 20 multiple-choice questions, each with 4 possible answers.
For someone who makes random guesses for all of the answers,
find the probability of passing if the minimum passing grade is 70 %.
~~~~~~~~~~~~~~~~~~~~~

To pass, 70% (or MORE) of 20 questions should be answered/guessed correctly.

70% of 20 means that 14 questions or more should be answered/guessed correctly.

In other words, at least 14 questions should be answered/guessed correctly.



The probability to guess correctly for each individual question is 1%2F4 = 0.25.
Guessing provides independent results for each of 2o questions.


So, we have the binomial distribution problem with 20 trials;
the probability of success is 0.25 for each trial.


I will use the MS EXCEL function BINOM.DIST.

Its format is  BINOM.DIST(k, n, p, L).

Here n is the number of trials (20); p is the probability of the individual success (0.25);
k the number of success; L is logical variable, pointing cumulative (if TRUE) or density function (if FALSE).


Using this function, I produced this table below


 k		 BINOM.DIST
------------------------------
 0              0.003171211939
 1		0.021141412926
 2		0.066947807600
 3		0.133895615199
 4		0.189685454866
 5		0.202331151857
 6		0.168609293214
 7		0.112406195476
 8		0.060886689216
 9		0.027060750763
10		0.009922275280
11		0.003006750085
12		0.000751687521
13		0.000154192312
14		0.000025698719    +
15		0.000003426496    +
16		0.000000356927    +
17		0.000000027994    +
18		0.000000001555    +
19		0.000000000055    +
20		0.000000000001    +


The column BINOM.DIST contains the values of probabilies (density function) for the values of successful trial from the first column.

What I need now is the sum of density values for k = 14, 15, 16, 17, 18, 19 and 20.

MS EXCEL allows to get it in couple of clicks.


The ANSWER, i.e. this sum, is about  0.00002951  (rounded).


ANSWER.  The probability to pass this exam is about 0.00002951.

Solved.

------------------

To learn more about function BINOM.DIST, look into this web-page

https://support.microsoft.com/en-us/office/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c#:~:text=Returns%20the%20individual%20term%20binomial,is%20constant%20throughout%20the%20experiment.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

As the other tutor points out, BinomDist is a great spreadsheet function to use.
You can generate a table as indicated and add up the values using the Sum command. Manually adding the values would be fairly tedious busy-work.

A quicker shortcut would be to use the cumulative flag of the BinomDist function.

The cumulative flag is the value of L at the end in the template
BinomDist(k, n, p, L)
If L = 0, then it computes an individual binomial value.
If L = 1, then it computes a cumulative sum

Typing in =BinomDist(13,20,0.25,1) will compute the binomial sum from k = 1 to k = 13.
The result of this calculation is approximately 0.999970488
Subtract this from 1 to determine the sum from k = 14 to k = 20.
1 - 0.999970488 = 0.000029512
Round this however instructed.

Side note: if your spreadsheet displays 2.9512E-05, or something along those lines, it means 2.9512 * 10^(-5). Then the -5 exponent moves the decimal point 5 spots to the left to arrive at 0.000029512