Binomial Distribution: Multiple Choice Test

[Monkeyseat]

An examination consists of 25 multiple choice questions each with five alternatives only one of which is correct. It is decided to subtract one mark for each incorrect answer and to award four marks for each correct answer.

a)

i) Find the mean mark for a candidate who guesses all the answers.

Mean = np
Mean = 25 * 0.2
Mean = 5

So they got 5 questions right. That means they got (5 * 4) marks for the questions they got right, but (20 * 1) marks have to be subtracted for the questions they got wrong. Therefore 0 marks?

ii) Find the mean mark for a candidate whose chance of correctly answering each question is 0.8.

Mean = np
Mean = 25 * 0.8
Mean = 20

So they got 20 questions right. That means they got (20 * 4) marks for the questions they got right, but (5 * 1) marks have to be subtracted for the questions they got wrong. Therefore 75 marks?

c)

For another candidate, the probability of correctly answering a question is 0.3. Find how many questions out of the 25 this candidate should answer to be 99.9% confident of answering at least one question correctly.

I don't know this and don't really have any working. I have been using Comulative Bindomial Distribution Function tables to answer some of these questions. For example:

http://courses.wcupa.edu/rbove/eco252/bintabl1.doc

I went down to the n = 25 table. I looked for the probability of 0.3 and went down the column until I saw 0.999. The first I saw was 15, so it would take them 15 questions? I think this is wrong though. I was told the answer was 20 but don't know why.

Please could you check (a) and (b) and also help me with (c).

Many thanks.

 

[tkhunny]

I like "a)i)". It is a rational design that creates an expectation of zero for a 100% guesser.

I do not like the terminology: "got ?? right". It is an expectation of getting that many correct answers, not a sure thing.

a)ii) looks as good as the first.

You may be a hair misguided on c).

I'd try this:

q = 1-p

\(\displaystyle P(All Wrong) = q^{n}\)

\(\displaystyle P(AtLeastOneCorrect) = 1-q^{n} \ge 0.999\)

 

[Monkeyseat]

I like "a)i)". It is a rational design that creates an expectation of zero for a 100% guesser.

I do not like the terminology: "got ?? right". It is an expectation of getting that many correct answers, not a sure thing.

a)ii) looks as good as the first.

You may be a hair misguided on c).

I'd try this:

q = 1-p

\(\displaystyle P(All Wrong) = q^{n}\)

\(\displaystyle P(AtLeastOneCorrect) = 1-q^{n} \ge 0.999\)

Thanks for replying.

Okay so P(Correct) = 0.3, therefore P(Incorrect) = 0.7.

P(AllWrong) = 0.7^25
P(AllWrong) = 0.0001341...

P(AtLeastOneCorrect) = 1 - 0.0001341... >= 0.999
P(AtLeastOneCorrect) = 0.999865893... >= 0.999

I understand that, but how do I figure out how many questions they have to answer? By the way I meant ">=" as greater than or equal to.

Thanks again.

 

[tkhunny]

You were not quite paying attention. Almost, but not quite.

Notice that I did not use '25'. Use my equation and solve for 'n'.

 

[Monkeyseat]

You were not quite paying attention. Almost, but not quite.

Notice that I did not use '25'. Use my equation and solve for 'n'.

Thanks again tkhunny. Here's what I've done (for simplicity I will use ">" rather than greater than or equal to):

1 - 0.7^n > 0.999
0.001 > 0.7^n
log 0.001 > log 0.7^n
log 0.001> nlog 0.7
(log 0.001)/(log 0.7) > n
19.367 > n

Since you cannot do .367 of a question, do you have to round it up to 20?

Thus, 20 > n. Is that correct? I thought it should be the other way around though, i.e. n > 20? I say that because, it would be better to answer all questions to ensure you get 1 correct.

I'm not sure if what I have done is right, this question is revision from an earlier section and comes before the section on logs!

 

[tkhunny]

Soooooooo...close!!

One tiny thing. You forgot your algebra!

log(0.7) < 0

What should do to an inequality when dividing by such a number?

Note: Don't forget your algebra.

 

[tkhunny]

I'm not sure if what I have done is right, this question is revision from an earlier section and comes before the section on logs!

You do not need the logarithms. You can just guess and check. n = 15? Not quite. n = 16? Still no...n = 19? So close! n = 20? Yes!!! No logs there.

Note: You'll look like a bigger genius if you use the logs.

 

[Monkeyseat]

Soooooooo...close!!

One tiny thing. You forgot your algebra!

log(0.7) < 0

What should do to an inequality when dividing by such a number?

Note: Don't forget your algebra.

Thanks again. Sorry, I know the sign is obviously the wrong way around, but I can't see which step you are pointing out where I made a mistake. :? I know when you divide by a negative number you flip the sign around. Please could you point out the mistake below:

1 - 0.7^n > 0.999
0.001 > 0.7^n
log 0.001 > log 0.7^n
log 0.001> nlog 0.7
(log 0.001)/(log 0.7) > n
19.367 > n

I'm not sure if what I have done is right, this question is revision from an earlier section and comes before the section on logs!

You do not need the logarithms. You can just guess and check. n = 15? Not quite. n = 16? Still no...n = 19? So close! n = 20? Yes!!! No logs there.

Note: You'll look like a bigger genius if you use the logs.

I was thinking trial and improvement by just substituting values in for n but I take it the method I adopted is acceptable. Thanks.

 

[Deleted member 4993]

Soooooooo...close!!

One tiny thing. You forgot your algebra!

log(0.7) < 0

What should do to an inequality when dividing by such a number?

Note: Don't forget your algebra.

Thanks again. Sorry, I know the sign is obviously the wrong way around, but I can't see which step you are pointing out where I made a mistake. :? I know when you divide and at which step are you dividing???by a negative number you flip the sign around. Please could you point out the mistake below:

1 - 0.7^n > 0.999
0.001 > 0.7^n
log 0.001 > log 0.7^n
log 0.001> nlog 0.7
(log 0.001)/(log 0.7) > n <<< right here it should be log(.001)/log(0.7) < n
19.367 > n

I'm not sure if what I have done is right, this question is revision from an earlier section and comes before the section on logs!

You do not need the logarithms. You can just guess and check. n = 15? Not quite. n = 16? Still no...n = 19? So close! n = 20? Yes!!! No logs there.

Note: You'll look like a bigger genius if you use the logs.

I was thinking trial and improvement by just substituting values in for n but I take it the method I adopted is acceptable. Thanks.

 

[Monkeyseat]

Thanks for the reply.

So when you divide by a log, do you always flip the sign over?

EDIT:

I know when you divide by a negative you flip the sign over, I didn't know you did it when you divide by a log.

 

[Deleted member 4993]

Thanks for the reply.

So when you divide by a log, do you always flip the sign over?

No!

log(0.7) = -0.15490196 = negative number

so when you divide by log(0.7) - you flip the inequality sign.

log(100) = 2 = positive number

so when you divide by log(100) - you do not flip the inequality sign.

 

[Monkeyseat]

Thanks for the reply.

So when you divide by a log, do you always flip the sign over?

No!

log(0.7) = -0.15490196 = negative number

so when you divide by log(0.7) - you flip the inequality sign.

log(100) = 2 = positive number

so when you divide by log(100) - you do not flip the inequality sign.

I see now! Many thanks. Now I know what tkhunny was getting at when he said "log(0.7) < 0" - I was looking for that step in my working and couldn't find it!

Thank you tkhunny and Subhotosh Khan for your help, very much appreciated.

By the way, was I correct in rounding 19.367 up to 20? It shouldn't be down to 19 as that would not give 99.9%?

 

[tkhunny]

Your answer is in your logic and in your expression.

1) You cannot answer 0.357 questions, so an integer is called for.
2) 20 is the integer that meets the requirements of the problem statement and the solution set.