Riddles, brain puzzles and mathematical problems

[Bridge]

It doesn't make any sense to subtract it though, because it's another possibility. In other words, there are three ways in which a current can pass through the circuit:

1) Via the top half (two switches)
2) Via the bottom half (three switches)
3) Via both (five switches)

Why is the solution not (2/3)^3+(2/3)^2+(2/3)^5? I honestly want to know, I have to take a final on this stuff next month and I find probability so vague and indirect.

[BAndrew]

It doesn't make any sense to subtract it though, because it's another possibility. In other words, there are three ways in which a current can pass through the circuit:

1) Via the top half (two switches)
2) Via the bottom half (three switches)
3) Via both (five switches)

Why is the solution not (2/3)^3+(2/3)^2+(2/3)^5? I honestly want to know, I have to take a final on this stuff next month and I find probability so vague and indirect.

As I said I haven't been taught about possibilities yet at school so I am not exactly sure, but I think
P(A U B) = P(A) + P(B) - P(A+B) *

*P(A+B) is symbolized with an inverted U

[Oscar House]

It doesn't make any sense to subtract it though, because it's another possibility. In other words, there are three ways in which a current can pass through the circuit:

1) Via the top half (two switches)
2) Via the bottom half (three switches)
3) Via both (five switches)

Why is the solution not (2/3)^3+(2/3)^2+(2/3)^5? I honestly want to know, I have to take a final on this stuff next month and I find probability so vague and indirect.

I think it has less to do with probability and more to do with the way electricity works. The current loops back to where it's coming from, causing a short circuit.

[Bridge]

It doesn't make any sense to subtract it though, because it's another possibility. In other words, there are three ways in which a current can pass through the circuit:

1) Via the top half (two switches)
2) Via the bottom half (three switches)
3) Via both (five switches)

Why is the solution not (2/3)^3+(2/3)^2+(2/3)^5? I honestly want to know, I have to take a final on this stuff next month and I find probability so vague and indirect.

I think it has less to do with probability and more to do with the way electricity works. The current loops back to where it's coming from, causing a short circuit.

EDIT: Never mind, what BAndrew posted makes sense (P(AuB) = P(A) + P(B) - P(AnB)).

[BAndrew]

Is the riddle solved or not? I am confused.

[Bridge]

Is the riddle solved or not? I am confused.

Yes, Chronofox said you were correct.

[BAndrew]

Riddle 3
Create an arithmetic representation that consists of exactly 6 eights (8), has no other numbers and the result is 1000.

The only symbols that may be used are those of addition, subtraction, multiplication, division, square root and brackets. You may also use expontation

In other words use
At least 6 eights -> 8 8 8 8 8 8
and with +,-,*,/,^,√,() try to make it equal to 1000.

It would be nice if someone could write this better so it is more understandable.

[Bridge]

(8^3+8^3-8-8-8)*8^0

Is this legal?

[BAndrew]

(8^3+8^3-8-8-8)*8^0

Is this legal?

No, you used the digit 0 and the digit 3. Use only eights.

8^8 for example is legal. 8^2 is not.

[Bridge]

(8^3+8^3-8-8-8)*8^0

Is this legal?

No, you used the digit 0 and the digit 3. Use only eights.

8^8 for example is legal. 8^2 is not.

Alright. You did say we were allowed to use exponents though.