probaility qns dse2013

[L14]

thanks @ron, kankgan arushi! I happend to see this after the war..lol..
someone enlighten us with qn 46's answer.
@kangkan, how did u get lamda=2 ?

[kangkan]

Hi Arushi..unfortunately it doesnt :) i am using rons method now

 possibilities [mm,ff,fm,mf]

let E be the event that- ff,mf,fm

Let A- be the event ff(this way the next letter will be from a female)

now we know that event A has happaned . ...and p(A)=3/4

now we need P(E/A).

So P(E/A)= P(E intersection A)/P(A)
              =P(FF)/P(A)=(1/4)  /(3/4)=1/3
so the mystery continues :)

[kangkan]

Hi Arushi..my sample was wrong :)

[kangkan]

Hi Ben.. the probabilty distribution of the number of failure(in this case no tips) is a possion(apparent from the answers)...now on everage,that is expected value of failrues is given as 2..for poisson distribution E(X)=lamda..hence here lamda is 2 :)

[The Villain]

@Kangkan ...how can probability of A be 3/4..
i know event A has taken place but 3/4....???

[kangkan]

sorry i mixed up event A and Event E

E- event that [ff,mf,fm]...p(E)=3/4

A:event -ff  p(A)=1/4

Now we need to find p(A/E)

=P(A inter E)/P(E)
=p(ff)/p(e)
=(1/4)/(3/4)
=1/3

[Arushi :))]

@kangkan
Yaar how can we take ''E- event that [ff,mf,fm]''
when we have to take E as the event that first letter is from female........so MF souldnt be there na.
E = ff, fm
Moreover,
I think the correct sample space is {FF,FM,MM}
Because its given that the postman has a ladies first policy in case there is a letter from female, so i think MF shouldn't be included....
I still think it should be 1/2.

[The Villain]

Even i think the ans is 1/2.
Either by taking sample {MM,FM,FF}
Or {MM,MF,FM,FF} ...but still i'm not too sure..

[kangkan]

hi Arushi even if its it is mf, the editor wil get the letter from the fmale 1st since the post man will deliver the letter from the girl first.. so if the letters are mf or fm, he will get f first

[kangkan]

Hi Arushi.i gtink this is fool proof

LetX be a random variable such X= no f letters from male correspondants. X can take values 2,1 or 0

X willl be a binomial distri with theta=1/2

Now P(X=0)=1/4

P(X=1) =1/2
P(X=2)=1/4

Now let E be the event that X= 1 or 0  P(E)=1/2+1/4=3/4
let A be the event that X= 0  therefore P(A)=1/4

Now P(A/E)= P(A inter E)/P(E)
               = P(X=0/P(X=1 or 0)
               =(1/4)/(3/4)
               =1/3
:)

[Arushi :))]

Pata nahi :(
Maybe its right,
even here P(X=1 ) consists of mf and fm.
What my approach is,
Sample space consists of all those possibilities in which the letters are received , and not delivered.
And we are doing the question from point of view of the receiver. So for me P(X=1) = 1/3

[L14]

I think kangkan is right. All four combinations {MM,MF,FM,FF} are possible with individual probabilities 1/4. Now if there is a letter from a female its always delivered first. So MF and FM imply the same.
The sample space is {MM,FM,FF} with probabilities {1/4,2/4,1/4}.
Event E-->The first letter is from a female.{FM,FF}. P(E)=3/4
Event A-->The second is also from a female.{FF}. P(A)=1/4
EnA--->FF p(EnA)=1/4
To find P(A|E)=p(EnA)/p(E)=(1/4)/(3/4)=1/3

[Arushi :))]

chalo maan liya
Thankyou kangkan, ron and ben :)

[kangkan]

[The Villain]

[Ashima]

Guys, I guess it's 1/3

See when the letters come in post office, they do not have anything to do with first female or not.

There is an equal probability of getting m or f

p(m)= 1/2
p(f)= 1/2


Now , total number of cases are [ mm, ff, mf, fm]

prob of mm, ff, fm and fm (each)= 1/2x1/2= 1/4

But for post office mf and fm are same, so we can consider hem as one event(X), and thus there prob will be 1/4+1/4= 1/2

Now, ( prob of 2nd is female| the first is from female) = prob (intersection)/ prob of first is from female

                                                                       = ff/(ff+(ff+X)
                                                                       = 1/4/ (1/4+1/2)
                                                                       =1/4/3/4= 1/3