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First i want to say there are a lot of questions related to this, but i couldn't find a similar case.

Suppose we have the typical problem where we need to compute the probability of pass a multiple-choice test. There are 8 questions with 5 options each, only one is correct. 4 of this questions are easy and the probability of do it well is $2/3$, the other 4 are hard and the probability of do it well is $1/5$. You can pass the test if answer correct at least 5 questions.

Im thinking in the binomial distribution here (from 5 to 8), but in this problem the probability of success is not the same for every question (like in similar problems). So maybe i should use the union of the probability of success for any type of answer (easy or hard), and then apply the binomial distribution.

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  • $\begingroup$ What exactly do you mean by "do it well"? Does it mean get the answer correct? Also, what is the significance of 5 options for each question? Are you implying the person taking the test can guess if they don't know the answer? $\endgroup$ – David Sep 26 '14 at 22:15
  • $\begingroup$ @David Yes, get the answer correct. Each question have 5 options to choose, only one is the correct. $\endgroup$ – Wyvern666 Sep 26 '14 at 23:15
  • $\begingroup$ Do the $2/3$rds and $1/5$th chances of getting the questions correct include the chance of randomly guessing the question correct? For example, on the hard question, you say $1/5$th chance of getting it correct. A random guess with no knowledge of the correct answer would give someone a $1/5$th chance of getting it correct. So do you mean the person has a $1/5$th chance of getting it right from knowledge alone? $\endgroup$ – David Sep 27 '14 at 12:30
  • $\begingroup$ There are 4 questions hard, and in this cases you chooses an option randomly, so the probability is 1/5, because you have 5 options. Anyway this probability is given in the problem, and the same for the probability to answer correct an easy question. In this other cases you have bigger probabilitys, 2/3, they are "easy". $\endgroup$ – Wyvern666 Sep 27 '14 at 16:20
  • $\begingroup$ For example the probability of get 4 hard ones correct, and the other 4 easy ones correct would be: $(1/5)^4*(2/3)^4*$ $\endgroup$ – Wyvern666 Sep 27 '14 at 16:21
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I think David's approach is correct.

This problem is about the sum of two independent binomial random variables.

Let $X$ be the number of easy questions answered correctly, and $P(X = x)$ the probability of getting x easy answers right. Let $Y$ be the number of hard questions answered correctly, and $P(Y = y)$ the probability of getting $y$ hard answers right.

Then define $Z = X + Y$, and let $P(Z = z)$ be the probability of getting $z$ answers right overall.

Then, under the assumption that $X$ and $Y$ are independent,
$$ P(Z = z) = \Sigma_{x=0}^4 P(X = x) * P(Y = z - x).$$

This is merely describing more formally how I suspect David arrived at his answers for $P(Z = 8)$ and $P(Z = 7)$.

Spelling out the method for $P(Z = 6)$:

$$P(Z = 6) = P(X = 4 \text{ and } Y = 2) + P(X = 3 \text{ and }Y = 3) + P(X = 2 \text{ and } Y = 4) $$

$$= (2/3)^4 6 (1/5)^2 (4/5)^2 + 4 (2/3)^3 (1/3) 4 (1/5)^3 (4/5) + 6 (2/3)^2 (1/3)^2 (1/5)^4. $$

Follow the same method for $P(Z=5)$. The probability of passing the test is:

$$P(pass) = P(Z=5) + P(Z=6) + P(Z=7) + P(Z=8).$$

I won't write down the whole calculation, but with the help of R (https://www.r-project.org), I worked out that the probability of passing this test is $0.196 $ (to $3$ d.p.) .

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  • $\begingroup$ Hey John, thanks for editing. This was my first post on StackExchange. Your edits will help me learn how to format equations better. Much appreciated! $\endgroup$ – Shabakuk Sep 29 '15 at 1:28
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I would study the different cases: how can you manage to get (at least) 5 correct answers?

you can:

1) Guess all 4 easy ones and don't miss all hard ones.

2) guess 3 easy ones and guess at least 2 hard ones,

3) guess 2 easy ones and at least 3 ones

4) guess only a easy one and all 4 hard ones

Note that these 4 different cases don't "overlap", therefore you can just sum up what you get from the single ones

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Break this up into cases:

All 8 answers correct: ($2/3)^4$ * $(1/5)^4$

Exactly 7 answers correct: [($2/3)^4$ * $(1/5)^3$ * $4$ * $(4/5)$] + [($2/3)^3$ * $(1/5)^4$ * $4$ * $(1/3)$]

Actually I will stop at this point because I am not even sure if this is right so far so I need someone else to chime in and assist/verify what is here so far is right or what I need to do to fix it.

If it is correct then the pattern for cases 6 and 5 should follow from cases 8 and 7.

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