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Wilson<3Pho
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I have a big question to ask about probabilities and estimated value.
The problem is there are 4 cups. In each cup there is 4 jellybeans. One green,one red, one blue, one orange. You have to pick one jellybean out of each cup. If you get 2 of the same, you get 50 dollars. If you get 3, 100 dollars. If you get all 4, you get 150 dollars.
So can someone help me with finding the probabilities?
 3 years ago
 3 years ago
Wilson<3Pho Group Title
I have a big question to ask about probabilities and estimated value. The problem is there are 4 cups. In each cup there is 4 jellybeans. One green,one red, one blue, one orange. You have to pick one jellybean out of each cup. If you get 2 of the same, you get 50 dollars. If you get 3, 100 dollars. If you get all 4, you get 150 dollars. So can someone help me with finding the probabilities?
 3 years ago
 3 years ago

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Owlfred Group TitleBest ResponseYou've already chosen the best response.0
Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
Ok, so it doesn't matter which one you pick from the first cup, but for all other cups they have to match it. So whichever one you pick in the first cup, what are the odds that you'll get the same from the second?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
err the probability rather. odds is something different.
 3 years ago

animalsavior94 Group TitleBest ResponseYou've already chosen the best response.0
i was thinking since a cup has 4 jelly beans and 1 each of a color so to get like example red u have a 1/4 chance to get it out of a cup...but idk..lol if it makes sense?
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
So to calculate the amount of probabilities, I have to find the chances of getting 1 1 0 0 then 1010 then so on and so on?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
No no no. You just have to find the probability for one cup. For example, if I pick a red bean first, what is the probability I'll get a red bean second?
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
but then i have to do it with each color :P so after wards calculating teh amount of getting 2s, i have to find 3s, 1110 1011 1101 and so on?
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
they are independent of each other the color of bean you get. The amount of money is not
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
Yes, but the probability will tell you what the expected earning is.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
So start with the first case. No matter which color you pick from the first cup, what is the probability you will pick that same color in the second cup?
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
1/4 Chance to get it from the first cup, but to get from second cup would i use a P(A and B) formula?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
What do you mean by a 1/4 chance for the first cup?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
The first cup doesn't matter. You can pick any color.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
You have a 1/4 chance for the second cup to match the first.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
and a 1/4 chance for the 3rd cup.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
and a 1/4 chance for the 4th cup.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
So P(A and B and C)
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
so that means to get 4 straight of the same color its a 1/64 chance?
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
Mhm. so if i wanted to find three of the same color, it would be P(A and B) and to find two just P(A) ?
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
So in an estimated value write out... 1/4(50) + 1/16(100) + 1/64(150) ?Or must I do this for every color
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
Nope, that's it. It doesn't matter which color you get first.
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
:0 no way, wouldnt i have to do 1/4(50) + 1/4(50) + 1/4(50) + 1/4(50) to represent the 4 different colors i might get?
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
And the nrepeat with the others?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
No. Because you'll only pick 1 of those 4 colors.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
And whichever one you pick you'll only have a 1/4 chance to get the same color on your next pick.
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
It seems too simple to be true :0
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
The expected earning is how much you will make on average per attempt.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
Each attempt you're only dealing with one color (the one you pick in the first cup).
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
Mathematics is often easier than you think.
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
Hmm, so if im trying to price this game like how much people have to pay to play
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
I just do the estimated value out
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
and just price on that number? So in this case its 21.1, so i might price it at 25?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
And add your profit margin.
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
Haha wow, in that case i think im good =D Thanks so much for your help! Ill post here if another problem arises :D
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
Will fan you! :D
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
be sure to click good answer to mark this question as solved.
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
Ohh mmkay no problem
 3 years ago

Wilson<3Pho Group TitleBest ResponseYou've already chosen the best response.0
Wait a minute, doesn't the probabilities have to add up to 1?
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.1
No they don't. The probability that you will pick a red, plus the probability to pick a green, plus the probability you will pick a blue, plus the probability you will pick an orange must add to one. But you don't care about those. You just want to know the probability you will pick the same color you picked from the first cup. If it added to 1 that would mean that you would always get the same color.
 3 years ago
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