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There are $8$ coins, $7$ of which have the same weight and the other one weighs more. In order to find the coin having more weight, a person randomly chooses two coins and puts one coin on each side of a common balance. If these two coins are found to have the same weight, the person then randomly chooses two more coins from the rest and follows the same method as before. The probability that the coin will be identified at the second draw.

my attempt $$[{(2/8)×(7/8)×(6/7)}+{(2/6)×(5/6)×(1/5)}] = 35/144$$

Please verify this. Thanks for reading.

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Probability that the heavy coin is not selected in the first draw is

$\frac{7}{8} \times \frac{6}{7} = \frac{42}{56} = \frac{3}{4}$

Probability that the heavy coin is selected in the second draw, given that it was not selected in the first draw, is:

$\frac{1}{6} + \frac{5}{6} \times \frac{1}{5} = \frac{2}{6} = \frac{1}{3}$

(because either the heavy coin is selected from remaining 6 coins as the first coin in the pair or it is selected from remaining 5 coins as the second coin in the pair).

So probability that the heavy coin will be identified on the second draw is:

$\frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$

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The heavier coin is identified on the second draw if the person first chooses two of the seven coins that have the same weight on the first draw and then chooses the heavier coin and another of the coins of the same weight on the second draw.

On the first draw, the person selects two of the eight coins. If the person does not choose the heavier coin on the first draw, the person must select two of the seven coins of the same weight. Thus, the probability that the person does not pick the heavier coin on the first draw is $$\frac{\dbinom{7}{2}}{\dbinom{8}{2}}$$ Assuming that occurs, there are six coins left. The person chooses two of them during the second draw. If the heavier coin is selected, then one of the $7 - 2 = 5$ remaining coins of the same weight must also be selected. Thus, the conditional probability that the heavier coin is selected on the second draw given that two coins of the same weight were selected on the first draw is $$\frac{\dbinom{1}{1}\dbinom{5}{1}}{\dbinom{6}{2}}$$ Hence, the probability that the heavier coin will be identified at the second draw is $$\frac{\dbinom{7}{2}}{\dbinom{8}{2}} \cdot \frac{\dbinom{1}{1}\dbinom{5}{1}}{\dbinom{6}{2}}$$

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