Moderator: Flannel Jesus
Since one of the $100 bills is already removed, then there is a 1 in 3 probablility of selecting the $1 bill and a 2 in 3 probablility of selecting another 100 dollar bill.
Flannel Jesus wrote:You got the right answer here, Motor, numerically. So congrats on that!
This might sound silly, but I don't know if the logic you used to get there makes sense though. I assume this bit is the meat of your reasoning:Since one of the $100 bills is already removed, then there is a 1 in 3 probablility of selecting the $1 bill and a 2 in 3 probablility of selecting another 100 dollar bill.
So I've devised an alternative scenario, to see if you can apply the same reasoning to this alternate scenario and still come to the correct conclusion.
This scenario is just like the first, but instead of 2 boxes, there are 3 boxes.
1 box has $1 and $100
1 box has $1 and $100
1 box has $100 and $100
Just like the first scenario, in this scenario you choose a box, essentially at random - you have no idea if you chose the box with 2x $100 or if you chose one of the 1 + 100 boxes. And once again, you pick out a bill at random, and you happen to find that you picked a $100.
So, just as in the first scenario, the question is, what' the probability that the other bill remaining in the box is also $100?
Ichthus77 wrote:Would you say the same thing about the boxes being a distraction if a 1 had been drawn?
Ichthus77 wrote:You’re not answering my question.
Ichthus77 wrote:You have 100% chance of drawing a $100 after a $1. The boxes are not a distraction.
It’s like my dad just let me win at chess. *eyeroll*
Ichthus77 wrote:The probability is the same in both cases.
Because in the first you have 4 total bills spread over two boxes, and in the second you have 6 spread over 3… 2 per box in each case… and same ratio of 1s to 100s.
Motor Daddy wrote:The boxes are a distraction, they mean nothing.
4 divided by 6 = .666... (66.666...%)
2 divided by 6 = .333... (33.333...%)
3 divided by 5 = .60 (60%)
2 divided by 5 = .40 (40%)
Flannel Jesus wrote:Ichthus77 wrote:The probability is the same in both cases.
Because in the first you have 4 total bills spread over two boxes, and in the second you have 6 spread over 3… 2 per box in each case… and same ratio of 1s to 100s.
Not the same ratio of 1s to 100s
Flannel Jesus wrote:The probability of pulling another 100 out of the same box in the second scenario is 1/2
obsrvr524 wrote:Flannel Jesus wrote:The probability of pulling another 100 out of the same box in the second scenario is 1/2
You lost me with that one.
Flannel Jesus wrote:How would you solve the problem?
Flannel Jesus wrote:But in simple terms,
- there are two ways that you could have chosen a box with $1 and $100, and
- also two ways you could have chosen the box with 2x $100.
Flannel Jesus wrote:You need to treat each $100 bill separately. So you could have chosen the box with 2x100 and chosen the first 100 bill, or you could have chosen the box with 2x100 and chosen the second 100 bill
Flannel Jesus wrote:I don't feel 100% comfortable with this wording, but it seems to consistently match the results that I get using Bayes theorem so I'll go with it:
You need to treat each $100 bill separately. So you could have chosen the box with 2x100 and chosen the first 100 bill, or you could have chosen the box with 2x100 and chosen the second 100 bill
Flannel Jesus wrote:Probabilities, statistics
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