DSE 2009 : Probablity

[sweta]

q10. (part A) A fair dice has numbers 1,2,3, 4, 5 and 6 on its sides. It is tossed once. I win Rs. 1
if an odd number shows up; otherwise I lose Rs.1.  Let X be the number that shows up
and Y the money I win. [Note: Y < 0 if I lose money.]  
Which of the following is incorrect?
a)  Prob (X > Y) = 1
b)  Prob(X= 3 | Y = 1) = 1/3
c)  E(Y) = 0
d)  Prob(Y = 1| X =5) = 1


q17) (Part 2)Jai and Vijay are taking a exam in statistics. The exam has only three grades A,
B and C. The probability that Jai gets a B is 0.3, the probability that Vijay gets a B is
0.4, the probability that neither gets an A, but at least one gets a B is 0.1. What is the
probability that neither gets a C but at least one gets a B?
a) 0.1
b) 0.6  
c) 0.8
d) Insufficient data to answer the question

[rahul]

option A : P(X>Y)=1 incorrect

[sweta]

hey can u tell me how did u ARRIVE ON IT?

[deepak]

For the first question,
P(X>Y) = Probability that the number on the die is greater than the amount won
          = 1 - P(number is less than or equal to winning)
Now, P(X>Y) = 1 only if the probability that the number on the die is less than or equal to winning, is zero. But this is not the case(when number on the die is 1, winning is also 1). So this is incorrect.

For the 2nd question, the answer is .6
To derive it, I would suggest this method. Visualize the problem like this :

So what is given is
P(no  A, 1 B) = P(J=B,V=B) + P(J=C,V=B) + P(J=B, V=C) = 0.1
Now you need to find
P(no C, 1 B) = P(J=B,V=B) + P(J=A,V=B) + P(J=B,V=A) and you will get this as 0.6

[s]

construct the prob distn table, P(X>Y)=1-P(X=1and Y=1)=5/6

look up Page 6 :
http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/notes_1.pdf

[sweta]

thank u deepak and s!