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You are in a dark room. There are 50 coins placed on table, out of which 10 coins are showing tails and 40 coins heads. Divide this set into 2 groups (not necessarily same size) such that both groups have same number of coin showing tails?

👁‍🗨 View Answer 👁‍🗨

Just pick 10 coins and 40 coins from the original 50 coins randomly and make 2 groups of 10 (group 1) and 40 (group 2) coins.
Now we just flip (upside down) the coins in the group of 10 and you get what you desire i.e same no of coins showing tails in both group.
How ?
You have groups of 40 and 10 coins. So the possibilities are
• Group 1- 10 Tails, Group 2 - 40 Heads. So when you flip the coins of Group 1, there will be 10 Heads and no Tail in Group 1 and 40 H and 0 T in Group 2. So 0 tails in both group..
• Group 1-9T,1H, ; Group2- 1T ,39H..Flipping Group 1-1T, 9H and Group 2- 1T and 39 H..So 1 tail in both groups.


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You are in a dark room. There are 50 coins placed on table, out of which 10 coins are showing tails and 40 coins heads. Divide this set into 2 groups (not necessarily same size) such that both groups have same number of coin showing tails?

👁‍🗨 View Answer 👁‍🗨

Just pick 10 coins and 40 coins from the original 50 coins randomly and make 2 groups of 10 (group 1) and 40 (group 2) coins.
Now we just flip (upside down) the coins in the group of 10 and you get what you desire i.e same no of coins showing tails in both group.
How ?
You have groups of 40 and 10 coins. So the possibilities are
• Group 1- 10 Tails, Group 2 - 40 Heads. So when you flip the coins of Group 1, there will be 10 Heads and no Tail in Group 1 and 40 H and 0 T in Group 2. So 0 tails in both group..
• Group 1-9T,1H, ; Group2- 1T ,39H..Flipping Group 1-1T, 9H and Group 2- 1T and 39 H..So 1 tail in both groups.


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@BrainyRiddles
@Curiosity_Tea

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