Discussion Problem_(13)

[duck]

Q1) A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Q2) Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

[Sumit]

Q1. Ans.2/5.
Q2. Ans.1/15.

[Mauli]

1)1 /6
2)1/15

[Sumit]

Q.A Man forgets his banker's card 10% of the time, he forgets his cheque book 5% of the time and he forgets both 2% of the time,
a)What is the probability that, on any one day, he will have both his banker's card and his cheque book?
b)What is the probability that he will find his banker's card in his pockets given that he has already found his cheque book?

[Mauli]

my ans
1) 87/100
2) 87/95.

[Sumit]

Why the answer of the first question posted by duck is 1/6????

[chiraxjr]

i think this 1/6 is wrong.

p(a/b)= 2/5///////5/36=1/18

[Sumit]

1/18 is also wrong...

[Sumit]

@mauli: me too getting
1) 87/100
2) 87/95.

[chiraxjr]

a=getting no 4 once
b=sum being 6
pa= 11/36
pb=5/36
pa&b=2/36
pa/b=pa&b/pb
=2/36/5/36=2/5

hope this ones rite!

[duck]

Hey Sumit.. it was mauli who posted those answers.. :)

[Mauli]

i m sorry
i forgot to exclude (4,4) which i hd been counted twice .
i would like to change my ans to the first ques to 2/11.
could someone pls cnfrm.

[Sumit]

@Duck; Yeah...I know,,just Forget to mention her name in that post...

[duck]

Correct Answers Sumit.. :)
1) 2/5
2) 1/15

[Mauli]

& how do we get 2/5?

[Mauli]

hi duck cn u pls tell wt is wrong with my approch
A=(1,5) , (2,4) ,(3,3) ,(4,2) ,(5,1)=5/36
B= Atleast one 4.
a= event when 4 is on the 1st dice=(4,1) ,(4,2) , (4,3) ,(4,4) (4,5) ,(4,6)
b= event ''    '' ''              2nd dice=(1,4) ,(2,4) ,(3,4) ,(4,4) ,(4,5) ,(4,6)
a intersection b= (4,4)
atleast 4 on one dice means=11/36=B
NOW , A intersection B is =2/36
so  P(A/B)= 2/11.
Where have i bungled up ?

[duck]

A= Sum of 6 is observed.
P(A)=5/36
B= Atleast 4 is observed.
A∩B = {(4,2) ; (2,4)}
P(A∩B) =2/36
Now, P(B|A) = P(A∩B)/P(A) =  2/5

[maahi]

how do u solve this ?

[Sumit]

As Duck did..

[Mauli]

thanks duck :)
i realise i was interpretng it as p(A/B).