Maths doubt

[Noel]

Thank you sir.

[Noel]

Q.3 Suppose X and Y are independent random variables with standard normal distributions.The probability of X<(-1) is some p€(0,1).What is the probability of the event:
X^2>1 and Y^3<(-1) ?
a. 3p
b. p^2
c. 2p^2
d. 3p^2

[Noel]

Q.4 Suppose n voters use the following procedure to find their leader.Each voter is given a fair coin.Each voter tosses his/her coin.A person is chosen leader if his/her toss outcome is different from that of the other n-1 persons' tosses.This procedure is iterated until a leader is determined.If n=3,then the probability of finding a leader in up to two iterations is
a. 7/16
b. 15/16
c. 39/64
d. 15/64

[Noel]

Q.5 There are three identical boxes,each with two drawers.Box A contains a gold coin in each drawer.Box B contains a silver coin in each drawer.Box C contains a gold coin in one drawer and a silver coin in another drawer.A box is chosen,a drawer is opened and a gold coin is found.What is the probability that the chosen box is C?
a. 2/3
b. 1/3
c. 1/2
d. 3/4

Q.6 A family has 3 children.Suppose the probability that a child will be a girl is 1/2 and that all births are independent.If the family has atleast one girl,what is the probability that the family has atleast one boy?
a. 5/7
b. 6/7
c. 5/6
d. 4/6

Q.7 Suppose player 1 has five coins and player 2 has four coins.Both players toss all their coins and observe the number that come up heads.Assuming all the coins are fair,what is the probability that player 1 obtains more heads than player 2?
a. 1/2
b. 4/9
c. 5/9
d. 4/5

Q.8 Let f(c)=max{x+2y|x>=0, y>=0, 2x+y=c}.The derivative of f at c is
a. c
b. 0
c. 2
d. c/2

[Amit Goyal]

Q 3
Pr (X^2>1, Y^3<(-1))
= Pr(X^2> 1)Pr(Y^3 < -1)   [Since X and Y are independent]
= [Pr(X < -1) + Pr(X > 1)] Pr(Y < -1)
= [p + p]p  [Since X and Y are random variables with standard normal distributions and Pr(X < -1) = p]
= 2p^2

[Amit Goyal]

Q.4
Let X be the number of iterations needed to find a leader.
Pr (X <= 2)
= Pr(X=1) + Pr(X = 2)
= 6(0.5)^3 + 2(0.5)^3[6(0.5)^3]
= 15/16

[Amit Goyal]

Q.5
Pr(Chosen Box is C| Gold Coin is found)
= Pr(Chosen Box is C, Gold Coin is found) /Pr(Gold coin is found)
= (1/6)/(1/2)
= 1/3

Q.6
Pr(Atleast one boy| Atleast one girl)
= Pr(Atleast one boy, Atleast one girl)/Pr(Atleast one girl)
= [1-Pr(no boy)-Pr(no girl)]/[1-Pr(no girl)]
= [6/8]/[7/8]
= 6/7

Q.7
Let X be the number of heads player 1 gets and Y be the number of heads player 2 gets. We want to find
Pr(X > Y)
= Pr(X > Y|Y = 0) Pr(Y = 0) +
Pr(X > Y|Y = 1) Pr(Y = 1) +
Pr(X > Y|Y = 2) Pr(Y = 2) +
Pr(X > Y|Y = 3) Pr(Y = 3) +
Pr(X > Y|Y = 4) Pr(Y = 4)

= Pr(X > 0|Y = 0) Pr(Y = 0) +
Pr(X > 1|Y = 1) Pr(Y = 1) +
Pr(X > 2|Y = 2) Pr(Y = 2) +
Pr(X > 3|Y = 3) Pr(Y = 3) +
Pr(X > 4|Y = 4) Pr(Y = 4)

= Pr(X > 0) Pr(Y = 0) +
Pr(X > 1) Pr(Y = 1) +
Pr(X > 2) Pr(Y = 2) +
Pr(X > 3) Pr(Y = 3) +
Pr(X > 4) Pr(Y = 4)           [Since X and Y are independent]

= (31/32)(1/16)
+ (26/32)(4/16)
+ (16/32)(6/16)
+ (6/32)(4/16)
+ (1/32)(1/16)

= (1/2)


There is another way to do the above problem. Since player 1 has five coins and player 2 has four, Player 1 will either have more number of heads than player 2 has or more number of tails than player 2 has but not both.
So, Pr(X > Y) + Pr(5 - X> 4 - Y) = 1
and by symmetry,
Pr(X > Y) = Pr(5 - X> 4 - Y)
Thus,
Pr(X > Y) = 1/2


Q.8
f(c)
= max{x+2y|x>=0, y>=0, 2x+y=c}
= max{z| c/2 <= z <= 2c}
= 2c
Thus, f'(c) = 2.

[Noel]

Thank you very much sir.

[Noel]

Q. An n-gon is a regular polygon with n equal sides.find the number of diagonals (edges of an n-gon are not considered as diagonals) of a 10-gon.
a. 20
b. 25
c. 35
d. 45

[Dreyfus]

Number of diagonals is given by n(n-3)/2

[L14]

A 10-gon has 10 points. Total number of lines that can be formed from 10 points are 10C2=45. But the sides are not diagonals. So total number of diagonals are 45-10=35.

[Noel]

Thank you vaibhav and ben10

[Noel]

Q. Consider the sytem of equations
Ax+By=0
Ux+Vy=0
A,B,U and V are i.i.d. random variables,each taking value 1 or 0 with equal probability.consider the following propositions.
(A) the probability that the system of equations has a unique solution is 3/8
(B) the probability that the system of equations has atleast one solution is 1.
a. Proposition A is correct but B is false
b. Proposition B is correct but A is false
c. Both propositions are correct
d. Both propositions are false

[Dreyfus]

These are homogeneous eq nd the possible solution if exist is Trivial solution i.e. x=y=0 now for solution to exist the determinant so formed must be not equal to 0.....ie AV-UB≠0
For this there will be 6 cases when D≠0 ...nd associated probability with each case is (1/2)^4
So total prob is 6*(1/2)^4=3/8 therefore proposition A is correct! Also homogenous equation are never inconsistent therefore either unique solution exists or infinitely many solution exists. So that prob of at least1 solution is 1.
 

[Noel]

Thanks vaibhav

[Noel]

Q. Two players,A and B, will play a best of seven tennis match (i.e.,the first to win 4 games will win the match, and the match will have atmost 7 games).The two players are equally likely to win any of the games in the match.The probability that the match will end in 6 games is
a. Less than the probability that it will end in 7 games
b. Equal to the probability that it will end in 7 games
c. Greater than the prob that it will end in 7 games
d. None of the above