A and B are two students. Their chances of solving a problem correctly are `1/3 and 1/4`, respectively If the probability of their making a common error is, 1/20 and the obtain the same answer, then the probability of their answer to be correct is

Options

A. `1/12`
B. `1/40`
C. `13/120`
D. `10/13`

Explanation

Let `E_(1)`= Event that both A and B solve the problem
`therefore P(E_(1))=1/3xx1/4=1/12`
Let `E_(2)`= Event that both A and B got incorrect solution of the problem
`thereforeP(E_(2))=2/3xx3/4=1/2`
Let E = Event that they got same answer
Here, `P(E//E_(1))=1,P(E//E_(2))=1/20`
`thereforeP(E_(1)//E)=(P(E_(1)capE))/(P(E))=(P(E_(1))cdotP(E//E_(1)))/(P(E_(1))cdotP(E//E_(1))+P(E_(2))P(E//E_(2)))`
`(1/12xx1)/(1/12xx1+1/2xx1/20)=(1//12)/((10+3)/120)=120/(12xx13)=10/13`