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?
If you were to play this game, the best combination would be if you had the 4 pennies and 2 of the gold coins in one cup and 1 gold coin in the other cup. You would have a 2/3 chance of winning 20 dollars. The worst possible combination would be if you had all the coins (pennies and gold coins) in one cup and the other one would be empty. You would have a 3/14 chance of winning.
ReplyDeleteThe worst combination Dan can make to win would be putting all the coins in 1 cup and leaving the other cup empty, this gives you a 3 in 14 chance of winning. The best combination to win would be to put a gold coin in one cup and put the rest of them in the other cup. This will give you a 2 in 3rd chance of winning.
ReplyDeleteConnor Mcquagge
The best choice for Dan is 1 gold in cup 1 and 2 golds and 4 pennies in cup 2 because the probability of winning is 2/3 chance, which equals 66%. The worst choice for Dan would be cup 1 empty and cup 2 full with 3 golds and 4 pennies, because the probability of winning is 3/14 chance, which is about .2% and that is the lowest probability out of all of the choices.
ReplyDeleteSydney McNabb
Testing the best way to get a gold coin is to have just one gold in the first cup and 2 others in the 2nd cup. This arrangement turns out to be 2/3 and i found this by adding up the fraction of that coin in the cups.
ReplyDeleteThe one that is the least way to combine the coins together is putting nothing in the first cup and then all of the rest ( 4 pennies, 3 golds), in the 2nd cup. This arrangement is 3/14 which isnt much of a chance. You had to do the same proccess you did to get the best combination as well.
The best arrangement for Dan and Ella to win in this game is container one g two gold and cup two containing four pennies and one gold coin, the probability for winning this combination is 2/3.As for the worst arrangement fro Dan and Ella I got the container with all the pennies and gold coins in one container and the other container being empty, the probability of winning this game would be 3
ReplyDeleteAfter finding all of the probabilities for Dan and Ella’s problem I found a best probability and a worst there were 10 different combinations and my best was the one that had one G in the 1st container and 4 P’s and 2 G’s because p(g) 2/12=1/6+1/2 then common denominators are 1/6+3/6=4/6=2/3 and 2/3 is the only one that is more than half. My worst one was the one with 2 P’s and 2 G’s in the 1st container and 2P’s and 1 G in the 2nd container because it equals 5/24 I found this by doing: P(g) 2/8+1/6= then common denominators are 6/24+4/24=10/48=5/24.
ReplyDeleteThe best combination is ppppgg/g because with the G being by itself makes it ½ of gold which gold wins. Having the G by itself makes it ½ which makes it the highest you can get without being empty. The fraction of winning is 2/3 winning gold.
ReplyDeleteThe worst combination is ppppggg/empty. The reason is because you have an empty cup and all the other coins (cup 1) will have a lower chance of being pulled. You have a 3/14 chance of pulling a gold from the cup. This makes that the lowest combination.
For this game the worst arrangement is the one with all the pennies and gold coins in one cup and the other cup empty. This is the worst arrangement because you only have a 3/14 chance of winning $20. The best combination to win this game is the combination with 4 pennies and 2 gold coins in one cup and 1 gold coin in the other cup. This is the best chance of winning $20 because you have a 2/3 chance of winning.
ReplyDeleteIn order to win this game you must draw a gold coin out of either cup. If Dan were to place a gold coin in one of the cups and the rest of the coins in the other cup, then he would have a 2/3 chance of winning. This arrangement would be the best choice. The worst choice would be to place all of the coins into one cup and leave the other cup empty, this would only give him a 1/14 chance of winning the game.
ReplyDeleteIf you wanted to win the game the bucket you wanted to choose is the one with 2/3’s.(this means that you would have 1 gold coin in one bucket and every other coin in the other bucket.) Because in the first container their was only one gold coin in the bucket. So this makes it very easy to chose a gold coin if you chose the first bucket.
ReplyDeleteIf you wanted to loose the game then you would choose the bucket that has every coin in it. This is true because the coins are not spread evenly. You would have a 3/14 chance of winning.
The worst combination is the cup with the empty side. Because the chance of picking a gold is 3/14. The best is the cup with gold in one cup and the rest in the other. Because the chance of picking the one gold is ½ and the rest is 1/12
ReplyDeleteThe best combination is if you had 4 pennies and 2 of the gold in one cup and 1 gold in the other cup. Other ways known as, G in container 1 and putting p,p,p,p,g,g and G in container 2 and that will give you 2/3. The worst combination Dan can make is putting all the coins in one cup and leaving the other one empty, so it would be p,p,p,p,g,g,g and container 1 is empty getting a total of 3/14.
ReplyDelete--Sydney Pickett
The worst combination for a chance of winning would be two browns in a cup along with 1 coin and another cup with 4 browns and 2 gold coins. I chose this probability because the greater chance of losing I split into half 2 browns and two browns the chance of getting a gold maybe small but the greater chance is spit but the probability of getting a gold coin is very small. The best combination is all browns in one cup and all gold in another because the chance of getting a gold is one half so you have that much of a chance to get a gold its much bigger than the probability we just discussed.
ReplyDeleteThe best combonation for getting a gold coin is putting G in the first container than putting P, P, P, P, G, and G in container 2. The worst combonation is having container 1 empty and have the rest of the coins in container 2.
ReplyDeleteMarissa Hackl
ReplyDeleteThe best arrangement for getting a gold coin is 2/3. This is because the gold on top is 1/2, and the two gold coins on the bottom are 1/12 each. When these are added together they make 2/3. The area model looks like G in the first cup then PPPPGG in the second. Since there is a gold coin in a cup by its self, it brings the chance of picking a gold coin up.
The arrangement that had the least probability of getting a gold coin is 3/14. This is the least because the first cup did not have any coins in it, and the second cup had all the rest of coins GGGPPPP. When they are all together the probability of getting a coin is 1/14 because there are 7 coins.
The best probability of getting a gold coin in Dan and Ella’s game is the arrangement PPPPGG/G with a probability of 2/3. The worst probability of getting a gold coin in Dan and Ella’s game is the arrangement PPPPGGG/EMPTY with a probability of 3/14.
ReplyDeleteThe best combination for putting these coins in a cup are if you put 1 gold coin in a cup and put 4 pennies and 1 gold coin in another cup. This would give you a 2/3 chance of winning. The least best combination for putting these coins in cup are if you put all the pennies and gold coins into one cup and left the other cup empty. This would leave you with a 3/14 chance of winning.
ReplyDelete-Christian Frantz
The best combination for Dan to pick would be PPPPGG/G. it would be this because .after I made the area model of each combination, I added up all of the different pieces and this combination gave me the highest percentage. The probability of Ella picking a gold coin using this combination is 2/3 and to find a percent u must divide the numerator by the denominator and 2 divided by 3 is .6 repeating which is the greatest percentage you can get with the following combinations. And the worst combination Ella could choose would be P/PPPGGG. Because after creating the area model and finding the probability it was 3/12. to find the probability of picking a gold coin you would need to divide the 3 by the 12 which would give you a percentage of 25. This is the lowest percent.
ReplyDeleteThe best arangement Dan can pick is having 1 gold coin in cup one and 4 pennies and 2 gold coins in cup 2. This is the best arangement because it has cup 1 which has a just the gold coin and that makes the probability goes up a bunch. The worst arangement for Dan to choose is 1 penny in cup 1 and three gold coins and 3 pennies in cup 2. This is the worst arangement because he has a higher chance of picking the penny in cup 1 and he wants the gold coins.
ReplyDeleteI think Dan should go with the combination of PPPPGG in the first cup and G in the second cup. I said go with that because it has a 2/3 possibility of getting a gold coin. In PPPPGG each one is worth 1/12 so the two Gs equals 2/12. In cup two G equals 6/12 since it is the only one there. The total you would get is 8/12 and if you reduce that you would get 2/3. The one you would have the least possibility on would be the PPPPGGG in one cup and empty in the other. Each coin would have 1/14 possibility of getting pick. Three gold coins so your possibility would be 3/14 and that is the lowest possibility there is for this game.
ReplyDelete