# Probability of passing a true-false test if a student knows some of the answers

A true-false test consits of 8 questions. A student will sit for the test, but will only be able to guess at each of the answers. [...] The following week, the same student will sit for another true-false test, this time there will be 12 questions on the test, of which he knows the answer to 4. What are the chances of passing this test (assuming that 50% is a pass)?

So my approach is that if he knows the answer to 4, then he only has to guess the answer to 2 or more of the remaining questions (that he has to guess answer to), because only then his score will be equal to or over 50%. So I would do:

If $X$ is the number of the remaining questions he answers correctly: $$P(X\ge2)= {8 \choose 2}0.5^20.5^6 + {8 \choose 3}0.5^30.5^7 + ...+{8 \choose 8}0.5^80.5^0$$

$$P(X\ge2)\approx0.9648$$

Is this correct? If no, what is the solution?

• What on earth does the first part of the question (the $8$-question test) has to do with all the rest??? Jan 6, 2016 at 10:58
• This task consisted of (a), (b), (c) and (d). The first sentence is an introduction to the first three, I included it in case someone will search for this question in the future. Jan 6, 2016 at 11:00
• BTW, the answer is simply $\sum\limits_{n=2}^{8} \binom{8}{n} \cdot \left(\frac12\right)^{n} \cdot \left(1-\frac12\right)^{8-n}$. Jan 6, 2016 at 11:01
• I think that when posting a question you should include only the relevant pieces of information. Jan 6, 2016 at 11:02
• Yes, it is. You have essentially split-ed (past-tense for "split") the scenario into disjoint events, and then added up their probabilities. I think that using the formula that I have specified in one of the comments above might emphasize this fact. Jan 6, 2016 at 11:04

$P(\text{Fail}) = {8 \choose 0} (\frac12)^8 + {8 \choose 1}(\frac12)^8 = \frac{1+8}{256}$