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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 $1/2$.

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P(both sides being white)$= 1/3$ (only 1 out of three has both sides black )

P(one side being white) $= 2/3$  ( two out of three has at least one side white)

P(both sides being white) / P(one side being white) $= \frac{1/3}{2/3} = 1/2$

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]

by

@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

correct
Pls correct me if i am wrong

Given - selected white face  = 1/3*1 + 1/3*1/2

To calculate other side probability is white is 1/2 => 1/3*1/2  /  (1/3*1 + 1/3*1/2)  = > 1/3
It is a case of conditional probability.

Probability of getting a white side = 3/6 = 1/2 (This is because we have total 6 faces. Out of them, 3 are white and 3 are black)

Given that one side is white, the probability that the other face is also white = (1/3) / (1/2) = 2/3