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?
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ReplyDeleteFor this game, the best combination to win is if one of the cups has 4 pennies and 2 gold coins and the other cup has 1 gold coin. This is the best chance of winning because you have a 2/3 chance of winning $20! The worst combination is if all the pennies and gold coins are in one cup and none in the other. This is the worst combination because you only have a 3/14 chance of winning $20.
ReplyDeleteThe best combination that Dan should choose is all the pennies and one gold coin in one container and two gold coins in the other container. The probability of getting a penny is 2/5, and the probability for getting a gold coin is 3/5.
ReplyDeleteThe worst combination is all the coins in one container and nothing in the other. The probability of getting a penny was 4/14 which reduces to 2/7, and the probability of getting a gold coin was 3/14.
The best one would be one cup with 1 gold and 0 pennies and the other cup with 2 gold coins and all of the pennies. The worst would be with one cup with nothing in it and the other cup with all the coins in it.
ReplyDeletebest choise is one gold in cup and 4 pennies and 2gold in the other it gives you a 2/3 chance.
ReplyDeletethe one that you have the worst chance in is all in one cup and none in the other you only have a 3/14 chance.
The best combination of coins in the two cups for Dan to choose would be to put all the gold coins in one cup and put all the pennies in the other cup. It is the best choice because Ella would have a 50% chance of getting a gold coin, since she will either pick a gold coin in one cup or a penny in the other. The worst combination of coins Dan could choose to put in the two cups would be putting one penny in a cup and then putting three pennies and 3 gold coins in the other cup. It is the worst choice because Ella would only have a 25% chance of grabbing a gold coin since in one cup there is only a penny and in the other there is an equal chance of grabbing a penny or a gold coin.
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ReplyDeleteThe arrangement with the 3/14ths probability is the worst choice possible for Dan because it significantly decreases his chances of winning. It has all of the pennies and gold coins in one cup, thus, making it harder for Ella to pick a gold coin. The arrangement with the 1/5 and ¼ probabilities is the best option. There are three pennies and two gold in one cup and one gold and one penny in another cup. This seems like it gives Ella a better chance at winning the game because she has an equal chance in one cup of getting a gold coin as she does for getting a penny, and close to an equal chance in the other cup with on three pennies and two gold coins.
ReplyDeleteThe best combination would be ppppgg in one cup and g in the other cup because you have 2/3 chance to win, the worst combination would be one penny in one cup and the rest in the other.
ReplyDeleteI think that for this situation the best chance would be to put ppgg in one cup and put ppg in the other cup. By doing this dan’s sister will have a 50 50 chance of picking the gold coin in one of the cups and winning $20 dollars. On the other hand the worst chance for them to win would be putting ggg and pppp in the other cup. I think this because if she chooses the wrong cup to choose from then they will not win anything. Also another chance that is bad is if you put ppppggg and no coins in the other cup. Just as same as the other chance, if you do this and Dan’s sister picks the wrong cup then they won’t be able to win anything,
ReplyDeleteThe best chance for Ella to pick a gold coin would be if all of the coins are in one cup with one gold coin in the other cup, I found the probability of picking a gold coin to be 2/3 which equals about 67%. The worst chance of picking a gold coin would be to have all of the coins in one cup and none in the other, which I found the probability to be 3/14 which is about 21%.
ReplyDeleteMost likely: The one with the most probability of picking a gold coin is where there is only 1 gold coin is in one cup and everything else in the other cup because there is a 1 half chance that he will chose the one with 1 gold and that makes the chances 1 half and there are 6 other coins in the other bucket and 2 of them are gold and 4 are pennies making a 1 twelfth chance for each of those coins making a 1 sixth chance of picking gold from that bucket and a 1 third chance of picking a penny added all together it is there is a 5 sixth chance of picking a gold coin and a 1 third chance of picking a penny
ReplyDeleteLeast likely: The one that is least likely to pick a gold coin is where there is only 1 penny in one container and everything else is in the other there s a 1 half chance of picking the container with a penny and that makes 1 half and in the other one there are 6 coins and there are 3 gold coins and 3 pennies each of them having a 1 twelfth chance of being chose and if you add up all the chances of picking gold in that one cup you get 1 fourth and if you add up all the chances of choosing a penny in that cup you get 1 fourth all the chances added you get 1 fourth chance of getting gold and a 3 fourth chance of getting a penny
By John Laurenza
ReplyDelete21/4/2011
The best combination is three Gold coins in one cup and four pennies in the other cup. This way you will have a 2/3 chances of getting three gold coins so that possibility is GG over PPPPG. There are ten possibility all to getter. So if he drew the three gold coins he would get $20. The worst chances is putting the gold and pennies to gather in one cup and the chances of getting a gold is 3/14 of a chances to get $20.
The best combination to choose would be one gold coin in the first cup, and four pennies with two gold coins in the other. The reason it would be the best id because you have 2/3, and a 67 percent chance of winning. The worst chance of winning would be Having all the coins in one cup, and none in the other, that would be a 3/14 which is a 21% chance of winning, therefore the first choice is the best answer out of all the other choices.
ReplyDeleteThe arrangement with the greatest possibility of winning is when you have four pennies and two gold coins in one cup, and one gold coin in the other cup. Because the fraction of winning is 2/3, which is the closest fraction to one whole. The worst arrangement of all the coins would be for all of the coins to be in on cup, and for the other cup to be empty. This is the worst arrangement because you only have a 3/14 chance of winning the twenty dollars.
ReplyDeleteThe best combination would be the cup with 4 pennies and 2 gold because it has the most posibility for it to win and what I had gotten for it was 4/3 because you would hsve to multiply 4/6 by 2/6 and get 4/3. The worst combination would be 1 gold and 1 penny because it not enough for someone to pick it, its half a chance that u would pick either 1 gold or 1 penny.
ReplyDeleteThe best combination would be in one cup having 2 gold coins and all 4 pennies. For the other cup put that one gold coin in there that you have left. That way if you chose that cup you have a perfect chance of winning. In the other it’s good because you have a 2/4 chance of winning. It still pretty fair. So whatever cup you pick you a good chance of winning. The worst combination would be in one cup have 2 pennies and 1 gold. Then in the other cup 2 pennies and 2 gold.
ReplyDeleteThe Best combination for Dan to choose to get a gold coin is if in cup 1 there is one gold coin and in cup 2 there has to be 4 pennies and 2 gold coins. With this combination you can get a 2/3 chance to get a gold coin which is 67%.this combination can get him $20. All other combination can get you a less probability to get a gold coin. The worst combination to get is if in cup 1 it’s empty and in cup 2 there are 4 pennies and 3 gold coins. This combination will get you a 3/14 chance to get a gold coin. With this combination, you can lose the game and not win the $20.
ReplyDeleteMrs. Ryon's 8th Grade Blog!: 7th period - Area Model Application Problem: "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..."
ReplyDeletethe best probability for this game would be the 3 greens and 3 purples in cup 1 and 1 purple in cup 2. the probability of this combanation is 3/6.
the wors combanation for this game was everything in cup 1 and nuthing in cup 2. the probability for this is 0/0.
The best combination is one gold in the first cup 2gold and 4 pennies in the second cup. He has a 67 percent chance of getting a gold coin. The worst combonation for ella is the first cup empty and the second cup with the three golds and 4 pennies he has a 3 and 14th chane of getting a gold
ReplyDeleteThe best combination for Dan to make is having four pennies and two gold coins in one cup and one gold coin in the other cup. The worst combination he could do is have an empty cup and four pennies three gold coins in the other.
ReplyDeleteThe best combination for Dan to choose would be the one with 2 gold coins and 4 pennies in one cup, and 1 gold coin in the other one. This is the best combination because, out of all the possible combinations, this has the greatest probability of getting a gold coin (the probability is 2/3 ). The worst combination that Dan could choose would be the one with 3 gold coins and 3 pennies in one cup and 1 penny in the other cup. This is the worst because it has the lowest probability of getting a gold coin, which is 1/4.
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