A test consists entirely of multiple choice questions, 27 questions in total. Each question has 5 options to choose from, of which only one is correct. Negative marking for incorrect answers does not apply and a mark of 50% for the test constitutes a pass.
(i) What is the probability that you will pass the test if you guess every single answer? Express your answer in percentage form, rounded to three decimal places.
(ii)Abby reckons she will increase her chances of passing by studying only one part of the work extremely well. Although only 11 of the questions were based on the work she studied, at least she is $100\%$ sure that she got all of them correct. If she guessed the remaining answers at random, what is the probability that she will pass the test?
Here's my attempt:
i) Making use of the binomial distribution,
$X \sim B(27,0.2)$
$p(x)= \frac{27!}{x!(27- x)!}(0.2)^x(0.8)^{27-x}$
$P[X \geq 14] = \sum_{x = 14}^{27} p(x)$
(to pass the test, you need to get at least $14$ questions correct)
ii) $X \sim B(16,0.2)$ ($27 - 11 = 16$)
$p(x)= \frac{16!}{x!(16-x)!}(0.2)^x(0.8)^{16-x}$
$P[X \geq 14] = \sum_{x = 14}^{27} p(x)$
Not sure if I am doing the right thing here. Please help.
Your answer to the first question is correct.
For the second question, since Abby knows the answers to $11$ of the $17$ questions and needs to answer at least $14$ correctly to pass the test, she needs to successfully guess at least $3$ of the other $16$ answers. If we let $Y$ denote the number of additional questions she must guess correctly, the probability that she passes the test if she knows $11$ of the answers is $$\Pr(Y \geq 3) = \sum_{k = 3}^{16} \binom{16}{k}(0.2)^k(0.8)^{16 - k}$$ That said, it would be easier to subtract the probability that she does not guess at least three of those $16$ answers correctly. $$\Pr(Y \geq 3) = 1 - \sum_{k = 0}^{2} \binom{16}{k}(0.2)^k(0.8)^{16 - k}$$