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?
29 comments:
The best combination was 4 pennies and two gold coins in cup one and one gold coin in the other. This is because the probability was 2/3 of a chance to win, which was the highest probability. For the worst combination I got all of the coins in cup one and none in cup two. The probability for this was 3/14, which was the least chance of winning.
The best combination is in cup 1 one gold and in cup 2 four pennies and two gold. This is the best combination because the probability of winning is 2/3 or two out of three.
The worst combination is in cup 1 three gold and four pennies and none in cup 2. This is the worst combination because the probability of winning 3/14 or 3 out of fourteen.
When Dan is arranging coins, the best combination he could make would be one gold coin in one cup, and the rest of the coins the second cup. Because he would have a 66% chance of winning. However, when Dan is arranging the coins, the worst combination he could have would be 1 empty cup, and all of the coins in another. Because then the chances of him winning is 21% with this combination.
The best choice for this problem would be putting one G in one cup and the rest in the other. This is like this because Dan will always have a G in one cup. This would lead to the probability of 2/3. The choices would be empty in one and all in the other, because you will have none in one cup and have more coins and pennies in that cup. Also the probability of this is 3/14.
The best combination for Dan to choose is one gold coin in one container and all the rest in another. This will give you a probability of 2/3, which is the highest probability that Dan could get. The lowest probability is having one container empty and the other container with all the rest of the coins in it. The probability of getting a gold coin is 3/14. This is least probability.
g/ggpppp= 2/3 is the largest because Dan will have the best chance to get a gold coin. Empty/gggpppp= 3/14 is the worst because he has the smallest chance of getting a gold coin. First, I found the probability then change all of them to decimals to compare them.
G/PPPPGG which equals to 2/3 is the greatest probability outcome of the experiment.
EMPTY/PPPPGGG which equals to 3/14 is the least probability outcome of this experiment.
To compare them all, I first found the outcomes in fractions, then changed them into decimals!
The BEST combination is two gold coins in one container with all of the pennies and only one gold coin in the other container without pennies. Which will get you get 2/3. The WORST combination is all the coins in one cup and the no coins in the other cup. This will get you a 3/14 chance at getting a gold coin.
Alexa nerren <3
The best situation would be 2/3. That would mean in container one you would have four pennies and two gold coins and in container two you would have just one gold coin. The worst would be 3/14. This would mean in container one all of the coins would be in there and in container two there would be no coins at all. I got this result because I used area models.
The best combination for Dan to win the $20 is to have all the gold coins in one cup and all the pennies in the other cup. The outcome in fractions would be 2/3 which is the greatest possibility to get the gold coins. The worst possible outcome for Dan would be all the coins in one cup and the other cup empty. The fraction outcome would be 3/14 which would be the worst outcome for Dan to choose.
In a game were the goal is to pick a gold coin from one of two cups the combination that will give you the best chance of winning is when there is one gold in one of the cups and two golds and four pennies in another. The worst combination is when there are no coins in one of the cups and all seven of the coins in the other cup. I got these answers by finding the probability for every possible combination then comparing them.
When you put 2 pennies and 1 gold coin in container one, then put 2 pennies and 2 gold coins in the other container you will have the HIGHEST probability. This is because 5/6 is almost a whole so that’s the highest. But the least likely is the one with no coins in one container and the rest of the coins in the other container. This is because the probability is only 3/14!!
The BEST choice is to do:
C1: G
C2: P, P, P, P, G, G
Because the probability for Green in this combo. Is 2/3, which is greater than any other combination.
The WORST choice is to do:
C1: EMPTY! :)
C2: P, P, P, P, G, G, G
Because the probability of green in this combo. Is 3/14, which is less than any other combination.
The best arrangement for Dan to choose when arranging the coins is one gold coin in one cup and two gold coins and four pennies in the second cup. The worst arrangement for Dan to choose when arranging the coins is four pennies and three gold coins in one cup and the second cup is empty.
If Dan and Ella want the highest probality, it would be PPPPGG in one cup and one G in the other. The probality would be 2/3, which has the highest probality. If she drew out of cup two, they are a 100 % chance of getting gold. The worst would be all them in cup one and none in the other. The probality for it is 3/14, is she picked the wrong one, it would be empty.
The best combination for Dan to choose a gold coin is a gold coin in one cup and the rest of the coins in another because the probability for a gold coin in this combination is 2/3. The worst combination is all the coins in one cup and the other cup is empty because the probability of getting a gold coin is 3/14.
The best combination for Dan to use is PPPPGG/G because the probability is 2/3. The wosrt combination for Dan to use is PPPPGGG/ empty because the probability is 3/14.
The best choice is isolating one gold coin in a bucket and putting the rest in the other bucket, which gives a 2 out of 3 chance of winning since half the probability is getting a gold coin(because of the isolated gold coin). The worst one is putting all the coins into one bucket and leaves the other one empty because it gives a 3 out of 14 chance and half the probability is eliminated.
The best winning combination for this game would be 1 gold coin in one cup and all the other coins in the rest. This is because if you draw out of the cup that the one gold coin is in, you have a 100% chance of winning, and even if you don’t you still have a chance of winning from the other cup. An area model also shows that the theoretical probability is the highest, at 4/6. The worst combination is every coin in one cup and none in the other. This is because there is a zero percent chance of winning on one cup, and there is a slim chance of drawing out of the other cup because there are more pennies than gold coins. An area model shows that the theoretical probability is 3/14, the lowest of the possible combinations.
The best combination for Dan to choose would be 1 gold coin in container 1 and the other coins on container 2. This would give Dan a 2/3 chance to win. The worst combination for Dan to choose is all the coins on one container. This would give Dan a 3/14 chance to win.
The best combination for Dan to choose when he is arranging the coins is 2/3. In cup one 4 pennies and 2 gold, in cup 2 one gold. This is because 1/2 times 1/6 is 1/12 and 1/2+2/12= 2/3. The worst possible combination is 3/14. the first cup empty and 4 pennies and 3 gold in the second cup. This is because 1/2 times 1/7= 1/14 so 1/14 three times is 3/14
the best combination is p,p,p,p,g,g in the first colum and g in the second colum. the probability is one-forth. the worst combination would be p,p,p,p,g in the first colum and g,g in the second colum. the probability was three-twentyiths. this is becasue if you make the fractions, then multiply them, then add them you get the probability.
-your favorite syudent, Ann Katheryn. <3 (:
The best combination for Dan to choose will be one gold coin in the first cup and two gold and four pennies in the second cup. This is the best combination for Dan to choose because this combination gives a gold coin the highest percentage to be pulled out than any other combination. The probability of gold being pulled out is 2/3.
The worst combination for Dan to choose is putting all the coins in one cup with none in the other. This is the worst combination for Dan to choose because this combination gives gold the lowest percentage of being pulled out than any other combination. The probability of gold being pulled out is 3/14.
The best combination leading to Dan winning $20 is PPPPGG/G. In fraction form that is 2/3. Combination PPPPGG/G is the best because it is the greatest outcome of winning $20. The worst combination leading to Dan not winning $20 is PPPPGGGG/empty. In fraction form that is 3/14. Combination PPPPGGGG/empty is the worst because it is the least outcome of winning $20.
Mackenzie Kirksey- After working out the area models, I concluded that the BEST choice for Dan would be 2/3. This is because there is the closest equivalency in each of the cups this way, between the 4 pennies and 3 gold coins. The WORST choice is 3/14, because there is the least equivalency of all the combinations. It basically gives one cup nothing and the other all of the coins.
The best combination was green in one container, and all the rest in the other. The probability of drawing gold was two thirds. This makes since because no matter what container you choose you have a good chance of drawing gold. The worst combination was all the coins in one container and the other container empty. The probability of the worst combination was three fourteenths, this makes sense because in the second container you have a zero percent chance of drawing gold.
The best combination for Dan to chose would be the 3 gold coins in 1 cup and the 4 pennies in the other cup. This is the best combination because this has the best probability of choosing a gold coin, ½ or 50%.
The worst combination is 2 gold coins in one cup and one gold coin and the 4 pennies in the other cup. This is the worst combination because the probability of choosing a gold coin is only 1/5 or 20%.
The best combinatin would be two gold coins in a cup and one gold coin and all pennies in the other cup. This is the best combination because you have a 50% 50% chance of picking a gold coin the worst combination is two gold coins in a cup and all the penies in another cup because the chance of picking a gold coin is 20%.
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