# A Simple probability question

Suppose you are correct 90% of the time.

But in a particular case of 30 questions, you only get 15 correct. What are the odds of that?

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The question can be answered if we model it by thinking of $30$ rolls of a $10$-sided die. That may not be a reasonable model, if many of the questions are about the $10\%$ that we do not know. And being aware that things are going rather badly may affect performance. – André Nicolas Oct 11 '12 at 18:42
You're probably intending for someone to tell you the probability, not the odds of that occurring. If you interpret 0.5 as a probability, that means 50% of the time. If you interpret 0.5 as odds, that is odds of 1:2, which means "twice as likely to happen than not," which translates to a probability of one third (33.333...%) that the event happens. – Firefeather Oct 11 '12 at 19:31
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Oct 14 '12 at 15:00

If we assume that the correctness of your answers to the questions are independent (getting one question correct does not affect your chances of getting any other question correct), then (as said in Alex's answer) we can use a Binomial probability distribution to get our answer.

Here's how it works: If the correctness of your answers to each question are independent, this implies (by the laws of probability) that the probability of getting one answer right and another answer wrong is equal to the product of the probabilities of getting each answer right individually. (If $A$ and $B$ are independent events, then $P(A \cap B) = P(A) P(B)$.) If you are correct 90% of the time, and each of the thirty questions are independent, then the probability of getting a specific question right is $90\%$, and the probability of getting a specific question wrong is $100\% - 90\% = 10\%$. So any single arrangement of 15 correct answers and 15 incorrect answer has a probability equal to the product of fifteen $90\%$'s and fifteen $10\%$'s; that is:

$$P(\textrm{a specific arrangement of 15 right and 15 wrong answers}) = (90\%)^{15}(10\%)^{15}=(0.9)^{15}(0.1)^{15}$$

However, there is more than one way to arrange 15 correct answers and 15 incorrect answers. (For example, instead of getting all the first fifteen right, you get the first fourteen right and the sixteenth right.) In fact, the number of arrangements of 15 correct and 15 incorrect answers out of 30 answers is $\binom{30}{15}$ (thirty choose fifteen), which equals $\frac{30!}{15! 15!} = 155\,117\,520$. Each of these arrangements is disjoint (meaning you can't get more than one arrangement at the same time), which implies (by the laws of probability) that the probability of one arrangement, or another, or another, or another (and so on until the last arrangement) is equal to the sum of the probabilities. (If $A$ and $B$ are disjoint events, then $P(A \cup B) = P(A) + P(B)$.) Each arrangement has the same probability of happening, so the probability of that one of those 15517520 arrangements occurring is the probability of a single arrangement, multiplied by 155117520. So, now, the probability of any of the 155117520 arrangements of 15 correct and 15 incorrect answers is: $$\binom{30}{15}(90\%)^{15}(10\%)^{15} = (155117520)(0.9)^{15}(0.1)^{15} = 3.193732180051435815048 \times 10^{-8} \approx 0.000000032$$

In general: The Binomial probability distributions make this a lot shorter because it puts all that stuff into one formula. The probability of getting $k$ successes (here, $k$ questions right) out of $n$ independent attempts (here, out of $n$ questions asked) with a true probability of success represented as $p$ (here, a $(100 \times p)\%$ chance of getting a question right) is:

$$\binom{n}{k} p^k (1-p)^{n-k} = \frac{n!}{k! (n-k)!} p^k (1-p)^{n-k}$$

Final note: The above demonstrates the probability of getting exactly 15 right and 15 wrong. The probability of getting 15 or less right and 15 or more wrong is a slightly higher. You can get this by using the above formula (again using $n = 30$ and $p = 0.9$) for $k=15$, $k=14$, $k=13$, and so on down to $k=0$, and adding all those probabilities together (because these are also disjoint events). This probability is $3.559479269444171697124 \times 10^{-8} \approx 0.000000036$.

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Thanks for your helpful explanation! It is easier to understand and I can see the multiple ways to look at the question. – Kevinsb Oct 11 '12 at 23:43
You're welcome! Be sure to upvote the question since you found it helpful. Thanks! – Firefeather Oct 12 '12 at 1:23
I tried to upvote, but I have insufficient reputation ;-( – Kevinsb Oct 12 '12 at 5:58
Perhaps anyone who happens along here can click instead because I cannot. Please?! In any case, I greatly appreciate your beautiful and thorough explanation. I was still confused by Alex's answer.. – Kevinsb Oct 12 '12 at 6:00
Ah! Of course! I forgot about that. If you feel inclined, you may mark my answer as accepted, but you don't have to. – Firefeather Oct 12 '12 at 14:33

Binomial distribution $B(30,0.9)$

$$\binom{30}{15}\left ( \frac{9}{10}\right )^{15}\left ( \frac{1}{10}\right )^{15}\approx3.19373 \cdot 10^{-8}$$

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Thanks for your answer @Alex I had to do 1 / 3.19*10^(-8) because I wanted it to be expressed that way? – Kevinsb Oct 11 '12 at 20:05
@Kevinsb, do you mean you want to express it as "a one-in-____ chance of getting 15 questions right"? – Firefeather Oct 11 '12 at 20:33
@Firefeather Yes. – Kevinsb Oct 11 '12 at 23:42
@Kevinsb, that would work; that works out to about 31311329, so you could say that there is about a 1 in 31,311,329 chance of getting 15 questions right. – Firefeather Oct 12 '12 at 1:21
@Firefeather Thanks! Yes, that seems like a helpful way to think about very long odds. – Kevinsb Oct 12 '12 at 5:55