Dan and Ella are paying $4 to play a game at the school carnival. Ella is blindfolded. Dan is given 7 coins to place in 2 different cups. He is given 4 pennies and 3 gold coins (identical in size). He randomly arranges them in the two cups in any combination he chooses. Ella is now instructed to choose a coin randomly out of one of the cups. If she chooses a gold coin, they win $20. If they choose a penny, they do not win anything. What is the probability that Ella will choose the gold coin in all possible situations ( I want a different probability for each situation)? Which combination would be BEST for Dan to choose when placing the coins in the cup? Which would be the worst?
18 comments:
The best chance of Dan and Ella winning $20, is if Dan puts one gold coin in one cup, and 2 gold coins, and four pennies in the other cup. This arrangement would give Dan and Ellen a 2/3 chance of winning $20.
The worst chance of Dan and Ella winning $20, is if Dan puts all the coins (3 gold coins, and 4 pennies) in one cup, and leaves another cup empty. This arrangement would give Dan and Ella a 3/14 chance of winning $20.
The best possible combination for Dan to place the coins in the cup is to put 4 pennies and 2 gold coins in cup 1 and to put 1 gold coin in cup 2. This would give Ella a 2/3 chance of pulling out a gold coin. The worst possible combination for placing the coins is to put all the coins in one cup and leave cup 2 empty. This combination would only give you a 3/14 probability of picking a gold coin.
The arrangement that will give you the greatest chance of winning is putting one gold coin in container one, and four pennies and two gold coins in container two. This arrangement’s probability of winning is two wins out of three. The arrangement that has the least chance of winning is no coins in container one, and three gold coins and four pennies in container two. This arrangement’s probability of winning is three wins out of fourteen. Dan and Ella should use the first arrangement to win the game.
The best combination that Dan could choose is one gold coin in cup 1 and the rest of the other coins in cup 2. The worst would be having one penny in the first cup and the rest in the second cup.
The best combo for Dan to choose is having Cup 1 holding 1 one gold coin, and Cup 2 housing the other coins. This is because there is now a hundred percent chance of picking a gold coin in Cup 1, which increases the overall probability of picking a gold coin. The worst combo is having cup 1 empty, and cup 2 housing the coins. This is because there is now only a three-fourteenths probability (or about 21%) of picking a gold coin. This is because there is no chance of getting a gold coin in the first cup.
The best possible choice for Dan and Ella is container 1:gold container 2:gold,gold,penny,penny,penny,penny. I say this because you have a two thirds chance of pulling out a gold coin from either container 1 or from container 2. the worst possible choice for Dan and Ella is container 1: empty container 2:gold,gold,gold,penny,penny,penny,penny. I say this because Dan and Ella will have a three fourteenths chance of pulling a gold coin out container 1 or container 2. The work I have to show for my statement was that I made 10 area models showing the possible ways of pulling a gold coin out of a container with three gold coins and 4 pennies.
If Dan puts 1 gold coin in one cup and the rest in the other cup Ella will have a better chance of choosing a gold coin, which is 2/3.If Dan puts all the coins in one cup and leaves the other one empty the chance of getting a gold coin will be 3/14.therefore the best chance of getting a gold coin will be 2/3 and the worst would be 3/14.
The best chance of Dan and Ella picking a gold coin would be if they put 1 gold coin in a cup and 2 gold coins, 4 pennies in the other cup. This would give them 2/3 probability of winning $20. The worst would be putting all coins in the same cup the probability is 3/14 to get $20. If he does this that would mean she has a 50% chance to get the cup with the coins and if she doesn’t there is no way for her to win. If I was Dan I would do the one with the best probability to win the $20.
Carlisle hattons blog btw.
On the worksheet that we were assigned we were suppose to find the best and worst probabitlity of getting the gold coin. In this I found that the best probability of getting a gold coin in this case would be in container 1 having 4 pennies and 2 gold coins, and in the second container having only 1 gold coin. When you do the area model for that problem you get the answer of 2/3. In this situation dealing with the opposite side of the problem the worst probability of getting a gold coin would be having all the coins in container 1 and none in container 2. Doing the area model the worst probability of getting a gold coin would be 3/14.
The BEST combination for Dan to choose is when he puts one gold coin in cup number 1 and puts 2 gold coins and 4 pennies in cup number two. The WORST combination for Dan to choose is when he puts 3 gold coins and 4 pennies in Cup number 1 and leaves cup number two empty. I found these answers by changing all the probabilities to percents, and found the biggest and lowest percent.
-BETHANY
The best probability of Dan and Ella winning $20 is if Dan puts one gold coin in cup one and four pennies and two gold coins in cup two. The reason why is because the probability is 2/3 which is the highest. The worst probability is if Dan puts all the coins in cup one while cup two is empty. The reason why is because the probability is only 3/14 which is very small. These are the best and worst probability of Dan and Ella winning 20 dollars.
Dan and Ella had several options to figure out the best and worst ways to be successful and win at the carnival game. In order to win the game, player one must hide all the coins, (3 gold coins, and 4 pennies) in two cups and player two must pick out one gold coin from whichever cup. The best combination to win the game would be having one gold coin in one cup and the remaining six coins in the other, (G vs. GGPPPP). They had a 2/3 chance of pulling out a gold coin, in that combination. The worst combination, with the least chance of the players winning would be having one cup completely empty and the other cup with the remaining 7 coins, (empty vs. GGGPPPP). This combination had a 3/14 chance of winning. In order to win, I would suggest going with the greatest chance of winning and having one gold coin in one of the cups and the others in the remaining cup.
When it comes to answering these questions you have to know all the possibilities. So if you have 3 gold coins and four pennies that have to go in to two cups you have ten possibilities that could get you the winning cup. However, out of all of those ten possibilities if you put one gold coin in one cup then put two gold coins and four pennies in another cup you have a 2/3 chance of winning, with two out of three being the highest chance of picking a gold coin and winning $20. Although, the lowest chance of winning the $20 is putting all of the coins into one cup which is just stupid because you only have a 3/14 chance of winning with an empty cup.
If Dan arranges the coins in the cups for Ella to pick, they have a better chance of winning if he puts them in the best order. In container one he should put gold,gold,penny. In container two he should put gold, penny,penny,penny. She will have 2/3 possibility of getting gold and winning $20 dollars. The worst chance of getting gold is having container one empty and putting all the coins in container two. His chances of winning are 3/7.
The BEST combination for Dan to put in the cups is one gold coin in cup 1 and 2 gold coins and 4 pennies in cup 2 it would give her a 2/3’s chance of winning. She is going to have a better chance of getting her $20 dollars. The WORST combination for Ella would be 3 gold coins and 4 pennies in cup 1 and cup two empty because she would have a bad chance at getting money and she would have a 3/14 chance of winning but that’s all. So it would give her a bad chance of getting $20 dollars.
-Madeline Finken
The best chance for Dan and Ella to win the $20 is to put 2 gold coins in one and 1 gold coin in the other with the pennies this would give them a 2/3 chance for winning
The worst chance would be to have 1 cup empty and have all the coins and pennies in the other this would give them a 3/14 chance of winning
The best combination dan could choose of 4 pennies and 3 gold coins in two sperate cups would be, all 4 pennies and 2 gold coins in cup 1, and 1 gold coin in cup 2. These combinations would give ella the best possible chance which is a two thirds chance of picking a gold coin and winning the $20. The worst possible combination and the combination which would give them the worst chance of winning the $20 is placing all of the coins in the first cup and to leave the second cup empty. Which would give them the worst chance of three fourteenths to win $20
The best combination for Dan to choose would be to put one gold coin in one bucket, then 2 gold coins and 4 pennies in the other bucket. The chace of winning would be 2/3 (6/12+2/12=8/12=2/3). The worst combination would be to put the 3 gold coins and 4 pennies into one bucket, then have the other bucket empty. The probability of winning would be 3/14.
-Blair <3
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