A Perfect Score.

lv100-pokelove

New member
Joined
Jul 7, 2016
Messages
2
A game is being played: there are twelve boxes into which a ball is thrown. The twelve boxes all are painted according to six different colours – orange, blue, red, yellow, green and violet; each colour owns two boxes, so there are two boxes for each of the six colours. The chance of missing when throwing the ball is 0 : the ball has equal chance, in any given throw, of landing in any of the twelve boxes – there is no bias towards any of the boxes; despite how accurate a contestant's aim may or may not be! (just as a fair die has equal chance of landing on any of its six faces).

Eight of the twelve boxes have had a large white X marked on them; the remaining four have been left unmodified. It is clear to the contestants which boxes have been marked with X: one yellow box; one red box; both of the blue boxes; both of the green boxes; one orange box; and one violet box – making a total of eight marked boxes, out of the twelve. The boxes marked with X score higher than the others, when a ball lands in them.

A contestant is given five tries at the game: she will have five balls to throw at the twelve boxes before her, one after another. NB: every time a ball lands in a certain colour, BOTH of the boxes belonging to that colour will be removed from the game, leaving two less boxes left to throw at with the contestants remaining balls. This means no two balls can land in any same colour, or same box.
So, on the contestant's first throw, she will have all twelve boxes – and all six colours – to throw at; but on her last, her fifth, throw, she will only have four boxes (belonging to only two colours) left to throw at.

What is the probability of her landing all five of her balls, one after another, in boxes marked with X.

Could order/permutations have something to do with it?
 
You too tired, Harry?

Capitals = them boxes with the big white X:
AA BB Cc Dd Ee Ff

1st throw: 8/12 success, or 4/12 you're out

Take over Harry...

Yeah that's the easy part!

2nd throw:
If first throw is A, then P(X) = 6/10
If first throw is B, then P(X) = 6/10
If first throw is C, then P(X) = 7/10
If first throw is D, then P(X) = 7/10
If first throw is E, then P(X) = 7/10
If first throw is F, then P(X) = 7/10

Your turn Menace...
 
You should find the total amount of combos, first. It seems like the case by case method of multiplying probabilities contains too many different scenarios.
 

Another approach (with thanks to my brilliant son):

X stands for getting an X (doh!)

O stands for getting an orange box.
B stands for getting a blue box, etc

P(X|O) = P(X|R) = P(X|Y) = P(X|V) = 0.5
P(X|B) = P(X|G) = 1

Since there are 5 balls thrown, there will be only one colour left over at the end.

If that colour left over is RED, then the colours that were hit were O, Y G, B and V ... ie 2 colours with double X, and 3 with single X.
Same applies if left-over colour is ORANGE, YELLOW and VIOLET.

If the colour left over is GREEN, then the colours that were hit were R, O, Y, B and V ... ie 1 colour with double X, and 4 with single X.
Same applies if left-over colour is BLUE.

So, 2/3 of the time we have 2 colours with double X and 3 colours with single X

and 1/3 of the time we have 1 colour with double X and 4 colours with single X.

Since all orders of colours are equally possible,

the probability of getting 5 Xs is \(\displaystyle \frac{2}{3}*(1)^2*(0.5)^3+\frac{1}{3}*(1)^1*(0.5)^4 = \frac{1}{12}+\frac{1}{48} = \frac{5}{48}\approx0.10417\)
 
Last edited:
I use UBasic.
As example, the command:
p = INT(RND*12)+1
calculates a random number from 1 to 12
and stores it under variable p.
Denis, you ought to look into P-BASIC. The P is for precision.
I have a textbook with a cd included. (Have no idea where it is)
It is a great tool for those of us who do like BASIC.
It really good for probability simulations.
 
Top