Question

there are three boxes
one has two red balls
another has two blue balls
third box has one red and one blue.

accordingly each box has label
RR - for both red
BB - for both blues
RB - for one Red & One Blue

Now some one changed the labels such that all labels are wrong now.

you can get one ball from any of the three box without looking in it.

how will you put back/identify correct labels?​

Answer 1

There are two boxes of RR and BB .

The other one is RB

After changing :

RB has been labelled something else .

This means that the new RB box has either both blues or both reds .

Open that RB box and draw a ball from it .

If you find red :

then change the label to RR .

The one with the BB level should be changed to RB .

The one with RR level should be changed to RB .

If you find blue :

then change it to BB .

The one with RR will become RB .

The one with BB should become RR.

Answer 2

Answer:

Let A be the event that you pick the box with two red balls,

let B be the event that you pick the box with two blue balls, and

let C the event that you pick the box with one of each.

Let R be the event that you choose a red ball.

Then P(A)=P(B)=P(C)=13 .

Also P(R|A)=1 , P(R|B)=0 , and P(R|C)=12 .

Also P(R)=1×13+0×13+12×13=12 (law of total probabilities).

Then P(A|R)=P(A)P(R|A)P(R)=23 ,

P(B|R)=P(B)P(R|B)P(R)=0 , and

P(C|R)=P(C)P(R|C)P(R)=13 .

I presume you want the conditional probability. The conditional probability that the other ball is red given that a red was drawn is 23×1+0×0+13×0=23 .

Of course the unconditional probability is 12 because picking one ball is the same as not picking the other. So if one is 12 , the other is 1−12=12 .