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Review of Key Theoretical Concepts Question 1 Suppose a standardized test contains 10 questions with four possible answers each. Assume the student guesses on each question by randomly picking an answer. What is the probability that a student gets the first 3 questions correct and the next 7 questions incorrect, given that he is guessing? Assuming that psychological factors do not play a role in solving an exam, then we can say that the probability of getting one question correct or incorrect does not depend on previous answers. Thus, the chance of getting one question correct is $P(C)=1/4$, and the probability of getting one question wrong is $P(I)=3/4$. Then the probability of getting first 3 questions correct and the next questions incorrect (let us call it $P(X)$ must be the same as: $$ P(X)=P(C)^3P(I)^7=(\frac{1}{4})^3(\frac{3}{4})^7=0.0020 $$ What is the probability that a student gets exactly 3 questions correct, given that he is guessing? Since, the probability of getting exactly 3 questions correct (let us call it $P(Y)$, implies getting 7 incorrect, but the order does not matter in this case, so we would need to know all the possible combinations of getting 3 right questions in the exam. To do so: $$ P(Y) = \dfrac{10!}{3!(10 - 3)!}P(C)^3P(I)^7 $$ $$ =\dfrac{10!}{3!(10 - 3)!}(\frac{1}{4})^3(\frac{3}{4})^7=0.25 $$ What is the probability that a student gets at least 3 questions correct, given that he is guessing? Being the probability of getting 3 or more questions correct $P(X\geq3)$, then: $$ P(X\geq3)=1-[P(X<3)]=1-[P(X=0)+P(X=1)+P(X=2)] $$ Fllowing the same logic as in the previous section, I am calculating one by one those probabilities:: $P(X=0)=\frac{10!}{0!10!}(\frac{1}{4})^0(\frac{3}{4})^{10}=0.0563$ $P(X=1)=\frac{10!}{1!9!}(\frac{1}{4})^1(\frac{3}{4})^9=0.1877$ $P(X=2)=\frac{10!}{2!8!}(\frac{1}{4})^2(\frac{3}{4})^8=0.2813$ With this now I can calculate $$ P(X\geq3)=1-[P(X<3)]=1-[(X=0)+(X=1)+P(X=2)]$$ $$=1-(0.0563+0.1877+0.2813)=0.4747$$ What number of questions correct has the greatest probability of occurring given the student is guessing (i.e., what is the mode)? For that, we have to calculate the probability for all the possible outcomes $P(X=0)$ to $P(X=10)$. $P(X=0)=\frac{10!}{0!10!}(\frac{1}{4})^0(\frac{3}{4})^{10}=0.0563$ $P(X=1)=\frac{10!}{1!9!}(\frac{1}{4})^1(\frac{3}{4})^9=0.1877$ $P(X=2)=\frac{10!}{2!8!}(\frac{1}{4})^2(\frac{3}{4})^8=0.2813$ $P(X=3)=\frac{10!}{3!7!}(\frac{1}{4})^3(\frac{3}{4})^7=0.2501$ $P(X=4)=\frac{10!}{4!6!}(\frac{1}{4})^4(\frac{3}{4})^6=0.1453$ $P(X=5)=\frac{10!}{5!5!}(\frac{1}{4})^5(\frac{3}{4})^5=0.0490$ $P(X=6)=\frac{10!}{6!4!}(\frac{1}{4})^6(\frac{3}{4})^4=0.0110$ $P(X=7)=\frac{10!}{7!3!}(\frac{1}{4})^7(\frac{3}{4})^3=0.0014$ $P(X=8)=\frac{10!}{8!2!}(\frac{1}{4})^8(\frac{3}{4})^2=0.0001$ $P(X=9)=\frac{10!}{9!1!}(\frac{1}{4})^9(\frac{3}{4})^1=0.0000$ $P(X=10)=\frac{10!}{10!0!}(\frac{1}{4})^10(\frac{3}{4})^0=0.0000$ The number of questions that has the greatest probability of occurring given that the student is guessing is 2. If a student actually scores the number correct identified in the previous part, should you infer he is guessing? The mode can give us a hint of the performance of the student in the exam. However, despite it could be likely that they are guessing, other factors could be into play: how many hours they slept before the exam, their average performance could be already bad, etc.