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There are five multiple choice questions on a test, with four choices per question. A student was given 10 questions to study for the test and the teacher picked 5 out of 10 questions to put on the test. The student memorizes 7 of the 10 answers of the questions given. If the student encounters the three questions the student does not know the answer to, the four choices will be equally likely to be guessed upon.

a. What is the probability the student gets the first question right?

b. What is the probability the student gets a 100 on the test?

I think I got the first part. Let A be the event the student gets the first question correct. For all questions to be rewarded a point, the probability is $\frac{7}{10}$ since she knows 7 of the the 10 questions. So we will call question one X1 and question two X2 and so on..

So we have $P(A)=P(A|X1=1)P(X1=1)+P(A|X1=0)P(X1=0)$=$(1)$($\frac{7}{10}$)+($\frac{1}{4}$)($\frac{3}{10}$)$=.7+.075=.775$

The second question I am more confused on. Let C1 mean question one is correct C2 question 2 is correct etc.

P(C1)=P(C2)=...=P(C5)=$\frac{31}{40}$

We will assume the C's are independent and do $(\frac{31}{40})^5$. After here I am not sure what to do and I am not sure if I am going on the right track. If somebody can guide me on the right track if the work looks good so far that be appreciated!

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  • $\begingroup$ "We will assume the $C$'s are independent" What? No, that is a bad assumption. Take for extreme example a two question test that the student studied only one question of and each question has a billion options. The probability of getting a particular question right is only slightly larger than $0.5$, and knowing that he got the first question right we can be almost certain that this was a result of it being the question he studied so the second question he is very likely to get wrong. $\endgroup$ – JMoravitz Feb 6 '20 at 2:39
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Your answer to the first question is correct, but your notation is poor. You say $X1$ denotes question one, so what does it mean to write $X1=1$? Instead I would define $B$ to be the event that question 1 is one of the 7 the student studied, and $B^C$ to be the event that question 1 is one of the 3 unknown questions. Then $$P(A)=P(A|B)P(B)+P(A|B^C)P(B^C) = 1 \cdot \frac{7}{10} + \frac{1}{4}\cdot \frac{3}{10} = 0.775.$$

For the second question, you're off to a good start, but we cannot assume the questions are independent. In fact, we know they are NOT independent. For instance, if the student gets questions 1, 2, and 3 all wrong, then those must have been the unknown questions and so they have to get questions 4 and 5 right.

Instead, let $E$ be the event that the student gets 100%. We can condition on the number of unkown questions $N$ the student sees. They either see $0$, $1$, $2$, or $3$ unknown questions, so: $$P(E)=P(E|N=0)P(N=0)+P(E|N=1)P(N=1)+P(E|N=2)P(N=2)+P(E|N=3)P(N=3).$$ So now you need to calculate $P(E|N=k)$ and $P(N=k)$ for $k=0,1,2,3$. As an example, I'll do $k=3$. Given that the student encounters $N=3$ unknown questions, the probability they get 100% is $1^2\left(\frac{1}{4}\right)^3 = \frac{1}{64}$. The probability they see $N=3$ unknown questions is $\frac{\binom{3}{3}\binom{7}{2} }{\binom{10}{5}}=\frac{21}{252}=\frac{1}{12}$ since there are $\binom{3}{3}$ ways for the teacher to choose the unknown questions, $\binom{7}{2}$ ways to choose the known questions, and $\binom{10}{5}$ total ways to choose 5 questions for the test. Can you do the rest now?

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  • $\begingroup$ I am not going to write out my full solution but basically let's say N=2 we had 7 choose 3 and 3 choose 2 over 10 choose 5 multiplied by $\frac{1}{4}$ squared and I did a similar thinking for each step and got $\frac{83075}{386304}$ $\endgroup$ – user710744 Feb 6 '20 at 3:12
  • $\begingroup$ Is that correct? $\endgroup$ – user710744 Feb 6 '20 at 3:13
  • $\begingroup$ Your reasoning looks correct. I haven't verified the final answer. $\endgroup$ – kccu Feb 6 '20 at 16:21
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Total ways of choosing the 5 questions is $\binom {10}5 $

To get 100% split into 4 cases

case 1: all five memorized , no lucky guesses

$$ P_1 = \frac {\binom {7}5}{\binom {10}5} $$

case 2: 4 memorized , 1 lucky guess

$$ P_2 = \frac {\binom {7}4 \times \frac14 \times \binom 31}{\binom {10}5} $$

case 3: 3 memorized , 2 lucky guesses

$$ P_3 = \frac {\binom {7}3 \times (\frac14 )^2\times \binom 32}{\binom {10}5} $$

case 4: 2 memorized , 3 lucky guess

$$ P_4 = \frac {\binom {7}2 \times (\frac14 )^3\times \binom 33}{\binom {10}5} $$

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