IIT JEE , AIEEE Forum - Mathematics , Algebra

neeraj_agarwal_1990

Ankur and ramesh have chances of solving a particular problem correctly are 1/8 and 1/12 respectively .they are given a question and they obtain the answer .If the probability of a common mistake is 1/1001 ,then the probability that the answer was correct is??
ans-13/14

i tried this way....
P(A)=1/8 P(R)=1/12 P(A' ^R')=1/1001 (^ = intersection) (' = compliment)

we have to find P(A^R) because according to the question, it seems as if they are getting the same answer....

P(A^R) = P(A)+P(R) - [1-P(A'^R')] but i m not getting the answer...

rathin_123

if A=ankur does right and R=ramesh does right

P(A)=1/8 P(R)=1/12

now both do same mistake P(A^R)' =1/1001 so P(A^R)=1-1/1001 = 1000/1001

so the probebility of correct answer is P(A' ^ R' )' [in ankur & ramesh any

one gets correct answer

=P(A)+P(B)-P(A^B)

=1/8+1/12-1000/1001

= -18995/24024 we will get it.we have

to find P(A' ^ R' )' ]

RyuAmakusa

the given ans is perfectly correct. I think that the q. is not worded properly.

is a simple application of bay's theorm.

1 1

---- * ----

8 12 13

ans= ---------------------------------------- = ---------

14

1 1 7 11 1

---- * ---- + ---- * ----- * ------

8 12 8 12 1001

ask if not understood.

neeraj_agarwal_1990

got it from somewhere else...neways...thanx...

rathin_123

RyuAmakusa plz post me the formula of bay's theorm.

RyuAmakusa

let a sample space(S) be divided into n events. E₁,E₂,.....E_n

E₁

E₂......E_n= S

E_i

E_j =

i, j

N

let A be an event in S

TOTAL PROBABLITY THEORM states that

P(A) = P(A/E₁) + P(A/E₂) + ....P(A/E_n)

WHERE

P(A/E₁) is the probablity of happening of A given that E₁ has happened

BAY'S THEORM STATES THAT

P(A/E_i) * P(E_i)

P(E_i/A) = --------------------------------------------

P(A/E₁) + P(A/E₂) + ....P(A/E_n)

ASK FOR CLARIFICATIONS IF ANY.

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