Discussion Problem_ (2)

[duck]

Q1) Find the probability that birthdays of six different persons will fall
in exactly 2 calender months?

Q2) Two fair dice are rolled. What is the conditional probability that the outcome is (1,1) given that 1 appears on one of the dice.

Q3) Suppose, we have 4 coins such that if the "i"th coin is flipped , heads will appear with probability "i/4" , i = 1, 2, 3, 4.
When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the second coin?

[Mauli]

thi is wht i am getting
1) 12C2*( 2^6-2)/ 12^6
2) 1/11
3) 1/5
please cnfrm.

[Sumit]

My take :
1).12C2*( 2^6)/ 12^6
2)1/10
3)If I'm not wrong this question is from Bayes theorem, still not able to find the Solution..
Plz provide the detailed Solution of it...
N also confirm other Sol....

[duck]

Hi.:)

Well done!

Correct Answers:
1) [12C2 (2^6 - 2)] / 12^6
2) 1/11
3) 1/5

 @Sumit:
3) It is to be done like this:
P(2nd coin|Heads) = [P(H|2nd coin)* P(second coin)]/P(H|1st coin)* P(first coin) + P(H|2nd coin)* P(second coin)+P(H|3rd coin)* P(third coin)] + P(H|4th coin)* P(4th coin)
= (2/4*1/4)/1/4*1/4+2/4*1/4+3/4*1/4+4/4*1/4
=(1/2)/1/4+2/4+3/4+4/4
=1/2 * 4/10
=1/5

[maahi]

@ duck : please explain how 1 is to b done.

[Sumit]

thanks a lot duck....actually in 1)I forget to subtract 2 cases...where all the 6 people has birthday in same month..
and in 3) actually I was using wrong formula....what a mistake..

n plz provide the sol. of 2nd as well....I'm still getting the same answer....

N plz keep posting this type of questions...we all appreciate it!!!

[duck]

@Maahi..

The number of ways in which you can select 2 months is 12C2.

Now, after selecting 2 months (say they are Jan and Feb), we want birthdays of six different persons to fall in exactly Jan and Feb.
Total number of ways in which birthdays of six persons can fall in Jan or Feb = 2^6.
(as each person has 2 options)
We know that there are 2 cases in which all birthdays will fall in exactly 1 month.
So, No. of favourable cases = 2^6 -2

Therefore, Probability= [12C2 (2^6 - 2)] / 12^6

[duck]

@Sumit:

Q2. Given (1,1)...(1,6),(2,1),(3,1),(4,1),(5,1),(6.1),
Prob of (1,1)= 1/11

[maahi]

thank you :-))

[maahi]

what two cases are there nd y r we dividing by 12 ^6??

[Sumit]

@thanks duck....

@Maahi: As duck said two cases will be birthday of 6 people falls into one month of two selected Months..

e.g suppose if we choose Jan n Feb From 12 months(i.e An element of 12C2)...we are subtracting two casses that is 1)all 6 people birthday fall in Jan
2) all 6 people birthday fall in feb..
we subtract it bcoz question says birthday fall in exactly 2 months...


-we are dividing by 12^6 bcoz.....these are the total no. of cases..(i.e no of ways birthday of 6 people can be distribute to 12 months, when each month can hv any no. of birthdays)..

[maahi]

got it thanks :-)

[Erika]

Hi duck

I did Q 3 differently although the ans is the same. I am putting it here so you could tell me if its wrong. In that case I would go with your method for any other similar ques. Tnx.

P(A=Head comes up) =1/4+ 2/4 + 3/4 + 4/4 = 10/4

P(B= 2nd coin was flipped) =1/4

P(A intersection B) = 2/4 (which is 2nd coin's prob of producing head)

So, reqd prob = 2 / 4
                        -----
                        10/4

=1/5

[duck]

Hey Erika.. :)

Thats not a correct way of doing it.
You've got P(A=head comes up) =10/4 > 1 which cant be possible.