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Suppose a box contains three cards, one with both sides white, one with both sides black, and one with one side white and the other side black. If you pick a card at random, and the side facing you is white, then the probability that the other side is white is $1/2$.

1 card both side white

1 card both side black

1 card 1 side white 1side black

from these 3 cards probability of choosing any one card is 1/3

Now while 1 side of the card is white , then the probability of other side also white

By Baye's theorem

1/3*P(W) /(1/3*P(W) +1/3 *P(B))

=(1/3*1/2) /(((1/3)*(1/2))+((1/3)*(1/2)))

=1/2 [Proved]

How's the Bayes formula applied here? - the terms are not clear.
No sir I think it is clear

Because we already know one side is white

Then probability of other side white

It is a conditional probability :)
yes, but that is intuition. We know only 2 coins are there with at least one face white. Try applying Bayes formula properly, you should not get 1/2 and this is a common mistake everyone makes in conditional probability.

@Srestha I think approach is correct. But wrong values substituted. Can u pls recheck once?

It is asked - Prob of other side white when one side is white

Nothing but using bayes theorem to identify White appeared what is probablity that source is unfair card with both sides white.

(1/3)(1) / (  (1/3)(1) + (1/3)(0) + (1/3)(1/2) ) = 2/3

@srestha

@Arjun Sir,

I think, the solution of @yg92 is correct.

just check it.

since in the question they have mentioned that one side of the card is white and they are willing to know or the favourable case is white on both the side.

Hence, the numerator should be (1/3) * 1, because 1/3 is for choosing the card and 1 as we know that its both faces are white, hence probability of white on that card will be 1 and that is what we want.

And the denomenator will the total probability.

@yg92

" the side facing you is white " is mentioned in question. Not both side should be white.

no where mentioned both side should be white