Question 1:
Answer: According to the Morgans Law
Question 2:
A)
0.0518x100=5.18%
B)
P(Two cards of the same rank)=1
Answer: the likelihood probability id 1. So 1 test is selected from each rank, 4 cards are
selected, 1 card is left, this card will be selected from any rank, so there will be at least 2
cards in 1 rank.
Question 3:
Question 4:
Question 5
Correct answered is let assign as B
And let C be the student has known the right answer
Let use Bayes Theorem
P(C/B)=P(B/C)xP(C)/P(B)
We know that P(C/B)=1
P(c)=p
P(B)=P(B/C)xP(C)+ P(B/C)xP(C)
P(B)=1 * p+!1/c *(1-p)
P(C/B)= p/p+(1-p/c)= c*p/c*p+1-p
Answer: The larger c there is a high probability that the student know answer correct
Question 6:
Test is first time positive is T1
Test second time positive T2
HIV is the virus
P(Hiv/T1T2)=P(T1T2/Hiv)*P(Hiv)/ P(T1T2)= P(T1T2/Hiv)*P(Hiv)/
P(T1T2/Hiv)*P(Hiv)+ P(T1T2/Hiv)*P(Hiv)=P(T1/Hiv)* P(T2/Hiv)*P(Hiv)/ P(T1/Hiv)*
P(T2/Hiv)*P(Hiv)+ P(T1/Hiv)* P(T2/Hiv)*P(Hiv)
P(Hiv/T1T2)=0.95*0.95*0.001/0.95*0.95*0.001+0.02*0.02*0.999=0.693
So there is a probability of 69.3% that after taking the test two times and this two times the
test is gonna be positive