probaility qns dse2013

[L14]

need help!
I think qn46 has something to do with poisson, but i don't get how to apply it.
and qn45 is it 1/4?
thanks.

[The Villain]

q46
@Ben ...I did it this way.


I get 2 letters per day.
Let M denote letter by male
n F denote letter by female
 so possible Outcomes {MM,FF,MF,FM}
I have already recieved the first letter which is by a female.
P(2nd is female | 1st is  female)
=(1/4)/(2/4)=1/2

[Arushi :))]

I agree with u on 45.
P( both first and second letter is from female ) =
P( first letter from female) * P ( second letter from female| first was from female)
= 1/2* 1/2
= 1/4
Is it right? :/

[Arushi :))]

good job ron.
Its 1/2

[The Villain]

Arushi..Is my intution right...Your one also makes
some sense bt m not too sure about your one

[Arushi :))]

Nai , yours is right,
I was finding it by intersection,
i.e both letters from female.
But the question is P( second F| first F)
so i am sure its 1/2 .:)

[The Villain]

Great then..
Wt about quest 46 ...ny idea...
i'm getting X~poiss(lambda=0)
i'm not able to carry on...

[kangkan]

i..i think the first answer is 4/3*e^(-2)... for poisson E(X) as N tends to infinity is lamda...here it is 2.

Sp P(X=3)= e^-2.(2)^3/3! hence the answer..i am not 100%sure

Seond one 2/3.

Let A-event that both letters were from female

B- one letter from female and one from male

E-event that 1st letter recived was from female

hence now the porb is P(A/E)..apply bayes theorem and ans comes out as 2/3 :)

[SoniaKapoor]

For quest 45 i think Ron is right.I'M geting 1/2 as well....and his explanation seems quite correct

[kangkan]

Soory its 1/3

P(E/A)=1 P(A)=1/4

P(E/B)=1 ,P(B)=1/2

P(A/E)= P(E/A)*P(A)/p(E/A)*p(A)+p(e/b0*pb
=1*1/4/ 1*1/4 +1*1/2=1/3

[kangkan]

Hi..sonia ..we forgot about the post man...the sample space (mm,ff,mf,fm) is equivalent to (mm,ff,fm) since the post man will deliver the letter by the female first always..

[SoniaKapoor]

I'm confused....guys please ask Amit Goyal Sir to upload the answers as he always does

[Arushi :))]

@ kangkan,
Yes, the sample space would be {MM,FF,FM}
So in such a case , if we use Ron's method,
P(second F| first F)= P( Both are female)/ P(first is female)= (1/3)/(2/3)= 1/2

               

[kangkan]

yes..i think that wud be best :)

[Arushi :))]

So, u agree with 1/2 ?

[kangkan]

Hi Arushi..i am not sure..i dont feel comfortable with intuition based solution..i feel better if its based on bayes theorem or someting like that..but i am not sure...

[kangkan]

i'll think tomo..gudnyt

[The Villain]

3 different methods and answers....pretty interesting question

[Arushi :))]

OKAY !! So i have solved the mystery :P It wasn't letting me sleep :D
Kangkan I am using Bayes theorem and it ill fetch the same answer.
S = { FF,MM,FM}
Let E = first letter from F = {FM,FF}
A= Both are from female = {FF}
B= First female and second male = {FM}
P(A)= 1/3
P(B)= 1/3
P(E/A) = 1, P (E/B)=1
P(A/E)= (1*1/3)/(1*1/3+1*1/3)= 1/2
So, Bayes Theorem gives the same answer on following Ron's method if we use new sample space by kangkan
So thanks Ron & kangkan :D

 

[The Villain]

Great job Arushi... :-)
so i guess this should bring an end to this "mast" question