Lottery

[SJ]

hey..can somebody pls help me figure out the solution of the lottery question (Q 35) in DSE 2010 paper.
The lottery is of the form (p1,p2,p3) where :
(0,1,0)>(0.1,0.89,0.01)
how does that help to infer which is preferred:
(0,0.11,0.89) or (0.1,0,0.9)

[s]

try this,

let u5 be utility from rs 5,u1 utility from rs1 and u0 utility frm rs 0

he prefers (0,1,0) to (.1,.89,.01) means Exp utility frm former is greater than exp utility of latter,so

0*u5+1*u1+0*u0>.1*u5+.89*u1+.01*01 =>u1>.1u5+.89u1+.01u0

u1-.89u1>.1u5+.01u0
.11u1>.1u5+.01u0
.11u1+.89u0>.1u5+.01u0+.89uo
.11u1+.89u0>.1u5+.9uo

so he prefers (0,.11,.89) to (.1,0,.9)

[SJ]

geez! thanks a great deal...
Can you also take a quick look at the q29 on conditional probability...plz...

[s]

P(x>=5 and x belongs to [3,8])/P(  x belongs to [3,8])=P(x belongs to [5,8])/P(  x belongs to [3,8])

k=1/25 so
find out P(x belongs to [5,8]): integrate over [5,8] using f(x)=1/25*(10-x) as its for range [5,10) ...its equal to 21/50

for P(  x belongs to [3,8])....integrate over [3,5] using f(x) = x/25 as its for range [0,5) and use f(x)=1/25*(10-x) for range [5,8] ...its equal to 37/50

so P(x belongs to [5,8])/P(  x belongs to [3,8])=21/50 /37/50 =21/37