Let's say that I had a test with three questions, and the probability of answering the questions right was .5, .4 and .33 respectively, presuming that the probabilities of each question are independent from each other. How do I calculate the probability of a) answering one question right, and b) answering two questions right? I remember about binomial distribution, but for that I need to have equal probabilities for each question, and I don't know how to solve this problem, so I would love the help of you guys. Thank you for the help
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3$\begingroup$ Shall we presume the three answers are independent? (That's rarely the case on any test, but it's often an implicit assumption in textbook questions.) Assuming this is a textbook question, then (1) please add the self-study tag to your post and (2) consider creating a table of all possible outcomes--there are only eight, so it's easy--along with their probabilities. It's the calculation of their probabilities that requires the independence assumption. $\endgroup$– whuber ♦Commented May 15, 2019 at 18:30
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$\begingroup$ It is independent, I forgot to say it in the question. So, I am trying to solve a more complex problem, in which are a lot more outcomes, but the calculation of those probabilities bugged me, then I tried to simplify it in the question. Is there some sort of equation that can calculate the probability instead of calculating all the possible outcomes? $\endgroup$– almeida07Commented May 15, 2019 at 18:38
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1$\begingroup$ Yes: see stats.stackexchange.com/questions/5347 for a discussion of those circumstances and stats.stackexchange.com/… for related threads. Because the probabilities differ and are (apparently) arbitrary, all such equations are equivalent to enumerating all the outcomes, which is why when the numbers get large, we have to approximate. It therefore would be better for you to ask about the problem you have rather than asking about an abstract one whose solutions might not help you -- or even mislead you. $\endgroup$– whuber ♦Commented May 15, 2019 at 18:47
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2 Answers
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With modest trickery one can pursue your binomial analogy by using simulation. Suppose there are three games $A, B, C$ with respective probabilities $0.5, 0.4, 0.33$ of winning and respective payoffs $1, 10, 100.$
Then in a 3-game series, you can find the probability winning (i) only the first game, game $A$, (ii) at least one game, or (iii) exactly one game, according as the payoff (i) equals $1,$ (ii) exceeds $0,$ or (iii) equals $1$ or $10$ or $100.$ Of course, there other possible events to explore.
A simulation in R of a million such series follows:
set.seed(1234) # for reproducibility
m = 10^6
a = rbinom(m, 1, .5); b = rbinom(m, 1, .4)
c = rbinom(m, 1, .33); x = a+10*b+100*c
mean(x > 0)
[1] 0.799284
mean(x == 1 | x == 10 | x == 100)
[1] 0.434121
TAB = table(x)/m; round(TAB, 3)
x
0 1 10 11 100 101 110 111
0.201 0.201 0.134 0.134 0.099 0.099 0.066 0.066
By simple probability rules:
(i) The exact probability of winning only the first game is $.5 -.2 - .165 +.066 = 0.201,$ which agrees with the simulated approximation to three places.
(ii) The probability of winning at least one game is $1 - .5*.6*.67 = 0.799,$ also simulated with three-place accuracy.
Notes: The eight probabilities in the summary table table at the end correspond one-to-one with the $2^3 =8$ regions of a Venn Diagram for this model--and with items comprising @whuber's recommended list. I do not claim that all of these probabilities are accurate to three places because I have not verified them all. [But with the same seed, a run with 100 million iterations (eventually) gave the same rounded values.]
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a) answering one question right (assuming one and only one)
this leads to P(answering one question right) = P(first right and last two wrong) + P(second right, first and last wrong) + P(last right and first two wrong)
the reasoning will be similar for b) but with all combinations that have two correct answers
You can also visualize this using a decision tree, and sum up all the combinations that satisfy the desired result
