Is this question an application of Baye's theorem or Bernoulli Trials? 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:


*

*Image of Baye's Theorem approach

*Image of Bernoulli Trials approach
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. 

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 $X\sim Bin(n,p)$.
