Key points
- There are eight coins, seven of which have the same weight and the other one weighs more.
- the coin will be
identified
at the second draw
Now how to identified in $2 nd $ draw?
In $1st $ draw , we must drawn $2$ unequal weight coin. Now, pick one coin from rest and put in common balance with any one of $2$ previous coin.
If it is equal weight, another coin of $1 st$ draw has more weight. This is only one possibility for it.
Now, maximum possibility will be $4.$
- Pick $1st$ two coin in common balance and got equal weight
- Pick next two coin in common balance and got equal weight
- Pick next two coin in common balance and got equal weight
- Now, pick only one coin and balance with any prev coin. If it has equal weight, then $8 th$ coin has more weight. Otherwise $7th$ one.
So, probability will be $\frac{1}{4}$