Maths

[Shweta Singhal]

A computer has 2 processors. The probability that processor one works is 3/4 & the probability that processor two works is 7/8. The computer will function if processor one works, or processor two works or both processors work. Find the probability that processor one is working, given that the computer is functional.

[duck]

Hi Shweta,

It can be solved in the following way>>

Let denote processor 1 by "I" and processor2 by "II" , Computer by "C"
P(I)=3/4    => P(I')=1/4
P(II)=7/8   => P(II')=1/8

NOW, P(C)=processor one works or processor two works or both processors work.
 
=>P(C)=3/4* 1/8 + 7/8 * 1/4 + 3/4 * 7/8
       
          =31/32
Also, P(computer is funtional given processor1 is working)= 3/4* 1/8 + 3/4*7/8
                                                                             =24/32

Therefore,

P(processor1 is working given computer is functional)= P(computer is funtional given processor1 is working)
                                                                        ---------------------------------------------------
                                                                                     P(C)

                                                          = 24/32
                                                           -------
                                                             31/32

                                                           = 24/32

[Amit Goyal]

Thats also correct Nidhi but there is a small typo towards the end. let me correct it on your behalf: P(I is working|C) = P(I is working, C)/P(C) = (24/32)/(31/32) = 24/31
What you computed as P(computer is funtional given processor1 is working) is actually not that but P(I is working, C).

[duck]

ohh.. Thank you sir..
...

[Shweta Singhal]

thank you sir...

[Shweta Singhal]

n thanks Nidhi..

[duck]

:)