Ihave a maths question and have no idea where to start: "john has an empty box. he puts some red counters and some blue counters into the box. the ratio of the number of red counters to blue counters is 1: 4 linda takes at random 2 counters from the box. the probability that she takes 2 red counters is 6/155 how many red counters did john put in the box? " i know that the probability of picking red the first time is 1/5, and therefore p(picking red the second time)=6/31(=(6/155)/(1/ and p(picking blue the first time is 4/5. but from here i don't know where to go or what to do.

X = number of red counters

ratio of reds to blues is 1:4
which can be thought of as x:4x when we multiply both sides of the ratio by x
meaning for every x red counters, there are 4x blue counters

We have x+4x = 5x total counters overall

The probability of pulling out a red is 1/5, so,
x/(5x) = 1/5

The probability of pulling out two reds is 6/155, so,
(x/(5x))*((x-1)/(5x-1)) = 6/155
Which is basically in the form A*B = 6/155
A = x/(5x) = probability of pulling out first red
B = (x-1)/(5x-1) = probability of pulling another red given the first is red; notice x drops to x-1, and 5x drops to 5x-1

Let's solve for x
(x/(5x))*((x-1)/(5x-1)) = 6/155
(x^2-x)/(25x^2-5x) = 6/155
155(x^2-x) = 6(25x^2-5x) ... cross multiply
155x^2-155x = 150x^2-30x
155x^2-155x - 150x^2+30x = 0
5x^2-125x = 0
5x(x - 25) = 0
5x = 0 or x-25 = 0
x = 0 or x = 25
Ignore x = 0 as its a trivial solution
The only practical solution here is x = 25

As a check:
A = x/(5x) = 25/(5*25) = 25/125 = 1/5
B = (x-1)/(5x-1) = (25-1)/(5*25-1) = 24/124 = 6/31
A*B = (1/5)*(6/31) = (1*6)/(5*31) = 6/155
So the answer is confirmed



There are 25 red counters

2160ft^3 i think you could use a formala to you out with this in the future