Suppose we have four chests each having two drawers. Chests 1 and 2 have a gold coin in one drawer and a silver coin in the other drawer. Chest 3 has two gold coins and chest 4 have two silver coins in each drawer. A chest is selected at random and a drawer opened. It is found to contain a gold coin. Find the probability that the other drawer has: (a) a silver coin, (b) a gold coin.
Answers
The probability that the other drawer has (a) a silver coin is and (b) a gold coin is .
- Given,
- There are 4 chests with 2 drawers each.
- Chest 1 and chest 2 contain one gold and one silver coin in their drawers.
- Chest 3 contain both gold coins in each of the drawer.
- Chest 4 contain both silver coins in each of the drawer.
- Since a chest is selected at random and found a gold coin.
- The probability that the other drawer has
- (a) a silver coin
- Since chests 3 and 4 contains both gold and silver coins respectively, these two get cancelled as they don't contain single gold and silver coins. We need to consider only 2 chests out of 4 chests, as chests 1 and 2 contain both one gold and one silver coins.
- Hence the probability becomes,
- (b) a gold coin
- Since one gold coin is already drawn, hence to find the probability that other coin is also a gold coin, we need to consider chest 3, as out of 4 chests, only chest 3 contains both the gold coins in it's drawer.
- Hence the probability becomes,
probability of Silver coin = 2/3 & probability of Gold coin = 1/3
Step-by-step explanation:
Chest 1 : Gold Silver
Chest 2 : Gold Silver
Chest 3 : Gold Gold
Chest 4 : Silver Silver
A chest is selected at random and a drawer opened. It is found to contain a gold coin
=> Drawer Selected in one of the First Three drawers
as drawer 4 does not have Gold coin
if Selected Drawer is Chest 1 & Chest 2
then other drawer will have Silver coin
if Selected Drawer is Chest 3 then other drawer will have gold coin
So
probability of Silver coin = 2/3
& probability of Gold coin = 1/3
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