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The question is as follows:

A student is given 8 true-false type questions. If he gets 6 or more items correct, he is declared pass. Given that he guesses the answer to each item, compute the probability that he will pass the test.

I've seen two different forums that approach this question using Baye's theorem and Bernoulli Trials and getting different probabilities. The images to replies on to these forums are:

I've also seen this question in two different textbooks (only giving the final result) also having taken these two very different approaches. Which of these two approaches are correct?

Thanks.

P.S The image of the Bernoulli Trials is just one many replies to the question posted on GMAT club forums.

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    Those links require some kind of signup. – 2017-02-26
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    Anyway, as I see it, it's definitely a "Bernoulli trials" type problem. – 2017-02-26
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    What are the final results at the links ? Beside this any reply is welcomed. – 2017-02-26
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    I'll add an image of the reply in those links. – 2017-02-26
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    And what do YOU think? – 2017-02-26
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    Bayes theorem approach: The definitions of the events don´t match the problem. This is the reason why the following approach doesn´t make any sense. – 2017-02-26
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    @Abhi2011 I´ve made an edit. Have a look to the remark. – 2017-02-26

1 Answers 1

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He passes the test if he guesses $6,7$ or $8$ answers right. The probability for that is

$P(X=6)+P(X=7)+P(X=8)$

$=\binom{8}{6} \cdot\left( \frac12\right)^6\cdot \left(\frac12\right)^2+\binom{8}{7} \cdot\left( \frac12\right)^7\cdot \left( \frac12\right)^1+\binom{8}{8} \cdot\left( \frac12\right)^8\cdot \left(\frac12\right)^0$

$=(28+8+1)\cdot \left( \frac12\right)^8=\frac{37}{256}\approx 14.45\%$

I don´t see why the Bayes theorem should be applied here.

Remark

Your approach works only because the outcomes are equally probable. The number of ways , disregarding order, that $k$ objects can be chosen from among n objects ist $\binom{n}k$. In your case $k$ are the right answers and $n-k$ are the wrong answers.

For $6\leq k \leq 8$ and $n=8$ we get the number of favorable outcomes

$\binom{8}6+\binom{8}7+\binom{8}8=28+8+1$

And the total number of possible outcomes is $\sum_{k=0}^8 \binom{8}k$. Here we can use the identity $\boxed{\sum_{k=0}^n \binom{n}k=2^k}$. Thus $\sum_{k=0}^8 \binom{8}k=2^8$

The ratio is $\frac{37}{256}$

But this approach wouldn´t work for $p\neq \frac12$. If you have this kind of exercises you should notice that the random variable is binomial distributed like $X\sim Bin(n,p)$.

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    By "your approach" I'm assuming you mean Bernoulli Trial's. Thanks for the help. :) Though, I still wonder why that textbook as well that forum would think of using Baye's Theorem. – 2017-02-26
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    @Abhi2011 Yes, I meant that link. You´re welcome. The most of us in the forum are wondering why the textbook has this solution. – 2017-02-26