Probability of passing an exam with different conditions

Question

An exam is scored out of 100 points and has a total of 10 questions. The minimum passing grade is 60 points. Each question is a TRUE/FALSE question. The first 8 questions are worth an equal number of points for a total of 60 points. The last two questions are each worth 20 points.

(a) Linus has decided to guess the answer to each question. He will not answer FALSE for any two consecutive questions. In how many ways can Linus answer all 10 questions?

(b) Patty has also decided to guess the answer to each question.

i. Given that Patty answers both of the last two questions correctly, what is the probability that she passes the exam?

ii. Find the probability that Patty passes the exam. [Hint: First condition on the number of correctly answered questions among the last two questions.]

So I'm not sure how to approach part a. I'm trying to minus the ways with 2 consecutive false from the total $2^{10}$, but don't know how to find the number of ways with 2 consecutive false. As for part ii of part b, I'm thinking that you use Bayes theorem?

Answer 1

Sum the contributions for every number of False answers from 0 to 5.

let $C_k$ be the number of strings of length 10 that consists of $k$ "F"s and $10-k$ "T"s and which do not contain the substring "FF"

either ...

the string ends in "T" , in which case the letter "F" will only appear in the combination "FT" . We will have to arrange $k$ substrings of "FT" with the remaining $10-2k$ "T"s

This can be done in $ \binom{10-k}{k} $ ways

or

the string ends in "F" , in which case the other nine positions must be filled by $k-1$ substrings of "FT" with the remaining $11-2k\;$ "T"s

This can be done in $ \binom{10-k}{k-1} $ ways provided that $k \ge 1$

so $$ C_k= \binom{10-k}{k} + \binom{10-k}{k-1}$$

so $$\begin{array}\\ N &=& \displaystyle\sum_{k=0}^5C_k \\&=& \binom{10}{0} + \binom{9}{0} + \binom{9}{1} + \binom{8}{1} + \binom{8}{2} ... \\&&+ \binom{7}{2} + \binom{7}{3} + \binom{6}{3} + \binom{6}{4} +\binom{5}{4} + \binom{5}{5} \end{array}$$

**** EDIT *****

for b) ii) ...

Sum the cases on the basis of how they did on the last two questions $$\begin{eqnarray*} P(\text {Pass}) =& P(\text{both correct}) \times P( \text{at least 20 pts. from first 8} ) \\+&P(\text{one correct}) \times P( \text{at least 40 pts. from first 8}) \\+&P(\text{none correct}) \times P( \text{at least 60 pts. from first 8}) \end{eqnarray*}$$

e.g. "at least 40 pts. from first 8" means you need to get at least 6 answers out of 8 (7.5 points per question) , so $$P( \text{at least 40 pts. from first 8})= \frac{\binom 86+\binom 87+\binom 88 }{2^8} $$