Suppose we have four boxes A,B,C and D containing coloured marbles as given below : One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? bo

Answer

N/a

Explanation

Let `E_(1)` is the event that randomly selected box is `A,E_(2)` is the event that randomly selected box is `B,E_(3)` is the event that randomly selected box is `C` and `E_(4)` is the event that randomly selected box is D.
Therefore events `E_(1),E_(2),E_(3)` and `E_(4)` are mutually exclusive and exhaustive.
`P(E_(1))=P(E_(2))=P(E_(3))=P(E_(4))=1/4`
Let `E` is the event that marble drawn is red.
`P(E/(E_(1)))=1/(1+6+3)=1/10`
`P(E/(E_(2)))=6/(6+2+2)=6/10`
`P(E/(E_(3)))=6/(8+1+1)=8/10`
and `p(E/(E_(4)))=0/(0+6+4)=0`
Using Bayes' theorem
`P` (marble drawn from box A) `=P((E_(1))/E)`
`=(P(E/(E_(1)))P(E_(1)))/(P(E/(E_(1)))P(E_(1))+P(E/(E_(2)))P(E_(2))+P(E/(E_(3)))P(E_(3))+P(E/(E_(4)))P(E_(4)))`
`=(1/10xx1/4)/(1/10xx1/4+6/10xx1/4+8/10xx1/4+0xx1/4)`
`=(1/4)/(1/4(1/10+6/10+8/10+0))=(1/40)/(1/4+15/10)=(1/40)/(15/40)=1/15`
`P` (marble drawn from box B)`=P((E_(2))/E)`
`=(P(E/(E_(2)))P(E_(2)))/(P(E/(E_(1)))P(E_(1))P(E/(E_(2)))P(E_(2))+P(E/(E_(3)))P(E_(3))+P(E/(E_(4)))P(E_(4)))`
`=(6/10=1/4)`
`=(1/10xx1/4+6/10xx1/4+8/10xx1/4+0xx1/4)`
`=(6/40)/(1/4(1/10+6/10+8/10+0))`
`=(6/40)/(1/4xx15/10)=(6/40)/(15/40)=6/10xx40/15=2/5`
`P` (marble drawn from box C)`=P((E_(3))/E)`
`=(P(E/(E_(3)))P(E_(3)))/(P(E/(E_(1)))P(E_(1))+P(E/(E_(2)))P(E_(2))+P(E/(E_(3)))P(E_(3))+P(E/(E_(4)))P(E_(4)))`
`=(8/10xx1/4)/(1/10x1/4+6/10xx1/4+8/10xx1/4+0xx1/4)`
`=(8/40)/(1/4(1/10+6/10+8/10+0))=(8/40)/(1/4xx15/10)`
`=8/40xx40/15=8/15`