\(P\left(C\ has\ the\ prize\mid B\ is\ open\ and\ empty\right)=\frac{P\left(B\ is\ open\ and\ empty\mid C\ has\ the\ prize\right)}{\left(P\left(B\ is\ open\ and\ empty\right)\right)}\ast P\left(C\ has\ the\ prize\right)\)
P(B is empty) can have 3 different cases as listed below
p(Neither B or C have the prize)=1/5
p( B does not have the prize but C Does)=2/5
p( B has the prize  and C does not=2/5
P(b is open and empty)=P(..|b has the prize but c does not)*P(B has the prize but c does not)+p(....|not b but c)*p(not b but c)+p...|neither b nor c)*p(neither b nor c)=\(\frac{1}{5}\ast0+\frac{2}{5}\ast\frac{1}{3}+\frac{1}{2}\ast\frac{2}{5}=0.333\)
Also we know that P(B is open)does not depend on P(B is empty)
\(\Rightarrow P\left(C\ has\ prize\text{| }B\ is\ opened\ and\ empty\right)\)\(\frac{P\left(B\ is\ opene\mid\ C\ has\ the\ prize\right)}{\frac{1}{3}}\ast P\left(B\ is\ emptye\mid C\ has\ the\ prize\right)\ast\frac{2}{5}\approx0.4\)
We know that \(P\left(A\ has\ prize\text{| }B\ is\ opened\ and\ empty\right)=P\left(D\ has\ prize\text{| }B\ is\ opened\ and\ empty\right)\) ...eq 1
\(P\left(A\ has\ prize\text{| }B\ is\ opened\ and\ empty\right)=\)\(P\left(B\ is\ opened\ and\ empty\text{| }A\ has\ the\ prize\right)\ast\frac{P\left(A\ has\ the\ prize\right)}{P\left(B\ is\ open\ and\ empty\right)}\)
    =\(P\left(B\ is\ opened\ \text{| }A\ has\ the\ prize\right)\ast\frac{\left(P\left(B\ is\ empty\ \text{| }A\ has\ the\ prize\right)\ast P\right)\left(A\ has\ the\ prize\right)}{P\left(B\ is\ open\ and\ empty\right)}\)
Similar to previous case we can find P(B opened and empty)=P(b is open|A has the prize)* p(A has the prize)=1/30
\(\therefore\ P\left(A\ has\ the\ prize\mid B\ is\ opened\ and\ empty\right)=\frac{1}{3}\ast\frac{1}{5}\ast\frac{30}{11}=0.181\approx0.182\)
from eq 1 \(P\left(D\ has\ prize\text{| }B\ is\ opened\ and\ empty\right)=0.182\)
Clearly Probability of the prize being behind C is still greater and therefore the contestant should not switch his choice and should stick with C.
(b) Prof. Chin knows that historically 2 out 75 students in his CS 542 class cheats (!) on his exams. Last semester, he suspected one of the students engaged in cheating during the final, but when he gently confronted the student, the student vehemently denied that he was cheating. If Prof. Chin has 90% accuracy in identifying cheaters (i.e. if a student is cheating, 90% of the time, Prof. Chin will indeed identify that the student as a cheater), but also has 20% false alarm rate (i.e. even though a student is not cheating, Prof. Chin will erroneously identify that student as a cheater). What is the probability that the student Prof. Chin confronted in the last semester’s final was indeed cheating?
P(student is really cheating|student is identified as cheater)=\(\frac{P\left(student\ is\ identified\ as\ cheater\mid student\ is\ really\ cheating\right)}{P\left(student\ is\ identified\ as\ a\ cheater\right)}\ast P\left(student\ is\ really\ cheating\right)\) ...eq1
Furthermore, P(student is identified as cheater)=\(P\left(student\ identified\ as\ cheater\mid student\ is\ really\ cheating\right)\cdot P\left(student\ is\ really\ cheating\right)+\)   \(P\left(student\ identified\ as\ cheater\mid student\ is\ not\ cheating\right)\cdot P\left(student\ is\ not\ cheating\right)\)=\(\frac{90\ast2}{100\ast75}+\frac{73}{75}\ast\frac{20}{100}=0.22\)    ..val 1
Using val 1 in eq 1
we get   P(student is really cheating | student is identified as cheater)= \(0.9\ast2\ast\frac{375}{75\ast82}=0.109\)
3. Linear Algebra
(a) Show that the inverse of a symmetric matrix is itself symmetric.
(b) Find the eigenvalues and eigenvectors for
4 .  Probability Distributions
2.2
The form of the Bernoulli distribution given by (2.2) is not symmetric between the two values of x. In some situations, it will be more convenient to use an equivalent formulation for which x ∈ {−1, 1}, in which case the distribution can be written
                                                     \(p\left(x\mid\mu\right)=\left(\frac{1-\mu}{2}\right)^{^{\frac{\left(1-x\right)}{2}}}\ \left(\frac{1+\mu}{2}\right)^{^{\frac{\left(1+x\right)}{2}}}\)