Topic: 0.30 btc bounty: maths help (statistics) - page 2. (Read 14215 times)
also why is this in development section?
Wording on this one is a little confusing. Don't let the product thing confuse you. You could just as easily say they each roll a die six times (if I am understanding the wording correctly) . and ask if they have matching numbers.
so for example your first "complementary pair" exercise would be translated from
"They have exactly 1 complimentary pair
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1)"
to
Alice rolled [1,2,3,4,5,6] Bob rolled [1,1,1,1,1,1,]
Since all numbers have an equal chance of being rolled, we can think of the problem as A) What are the chances that Alice and Bob each rolled one 1, B) what are the chances that Alice and Bob each rolled 2 ones, C) What are the chances that Alice and Bob each rolled three ones, etc
I don't want to do your homework for ya, even for money (someone else might though!) but this might help you on your way
so for example your first "complementary pair" exercise would be translated from
"They have exactly 1 complimentary pair
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1)"
to
Alice rolled [1,2,3,4,5,6] Bob rolled [1,1,1,1,1,1,]
Since all numbers have an equal chance of being rolled, we can think of the problem as A) What are the chances that Alice and Bob each rolled one 1, B) what are the chances that Alice and Bob each rolled 2 ones, C) What are the chances that Alice and Bob each rolled three ones, etc
I don't want to do your homework for ya, even for money (someone else might though!) but this might help you on your way
In question 6, for example, would lists A1,A1,A1,A1,A1,A1,B1,B1,B1,B1,B1,B1 be equivalent to A1,A2,A3,A4,A5,A6,B1,B2,B3,B4,B5,B6 for your purposes?
If so, each party will pick one of 6^6 lists, and the other person's choices don't matter, so the chances are 1 in 6^6.
Moving on to question 5, there are 6^1 ways to pick a list that matches in at least 5 places, but one of them is the answer to #6, so the chances are 1 in 6^5 minus 1/6^6.
In question 4, there are 6^2 ways to pick a list that matches in at least 4 places, but some match more, so we have to remove them again. So, 1 in 6^4 minus 1/6^5 minus 1/6^6.
Question 3, 6^3 possible, minus the extras, so: 1 in 6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Q2: 1 in 6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Q1: 1 in 6^1 minus 1/6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
If so, each party will pick one of 6^6 lists, and the other person's choices don't matter, so the chances are 1 in 6^6.
Moving on to question 5, there are 6^1 ways to pick a list that matches in at least 5 places, but one of them is the answer to #6, so the chances are 1 in 6^5 minus 1/6^6.
In question 4, there are 6^2 ways to pick a list that matches in at least 4 places, but some match more, so we have to remove them again. So, 1 in 6^4 minus 1/6^5 minus 1/6^6.
Question 3, 6^3 possible, minus the extras, so: 1 in 6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Q2: 1 in 6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Q1: 1 in 6^1 minus 1/6^2 minus 1/6^3 minus 1/6^4 minus 1/6^5 minus 1/6^6.
Fact 1:
Alice chooses 6 items from the following list of products:
A1, A2, A3, A4, A5, A6
Fact 2:
Bob chooses 6 items from the following list of products:
B1, B2, B3, B4, B5, B6
Fact 3:
They are free to choose the same product more than once, so for example, Alice might choose to buy the following 6 items:
A1, A1, A2, A2, A2, A6
Fact 4:
Each 'A' product is used with a corresponding 'B' product, so for example, an A5 needs exactly one B5 to be of any use, and a B5 needs exactly one A5 to be useful (Let's call this match a 'complimentary pair').
Questions:
From Bob and Alice's 12 items, what are the chances that:
For each correct answer with detailed workings I will give 0.05 btc (~$1) for a total of 0.30 btc (~$6).
Alice chooses 6 items from the following list of products:
A1, A2, A3, A4, A5, A6
Fact 2:
Bob chooses 6 items from the following list of products:
B1, B2, B3, B4, B5, B6
Fact 3:
They are free to choose the same product more than once, so for example, Alice might choose to buy the following 6 items:
A1, A1, A2, A2, A2, A6
Fact 4:
Each 'A' product is used with a corresponding 'B' product, so for example, an A5 needs exactly one B5 to be of any use, and a B5 needs exactly one A5 to be useful (Let's call this match a 'complimentary pair').
Questions:
From Bob and Alice's 12 items, what are the chances that:
- They have exactly 1 complimentary pair
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1) - They have exactly 2 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B1, B1, B1, B1) - They have exactly 3 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B1, B2, B3) - They have exactly 4 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B4, B4) - They have exactly 5 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B2) - They have exactly 6 complimentary pairs
(eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B6)
For each correct answer with detailed workings I will give 0.05 btc (~$1) for a total of 0.30 btc (~$6).
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