There are 45 discs y or b disks in a bag. If the probability of removing 2 y disks at random is 1/15, how many y discs are present initially?
A bag has 45 discs.
Let the number of y ( yellow discs ) be x.
Then the probability of choosing a y is P(y) = x/45
And the probability of choosing a b is P(b) = (45-x)/45 = 1- p(y).
If two were choosing randomly without replacement.
Then the probability if choosing the 2nd yellow = (x-1)/44
Given that the probability of getting 2 y =1/15.
But we know that the probability of choosing 2 yellow = probability of choosing the first yellow * probability of choosing the 2nd yellow.
P(2 y) = P (1st y) * P( 2nd y)
= x/45 * (x-1)/ 44 = 1/15
==> x(x-1) / 45*44 = 1/15
Let us cross multiply.
==> x(x-1) = 45*44/15 = 132
==> x^2 -x -132 = 0
Now we will factor the quadratic equation.
==> (x-12)(x+11) = 0
==> x1= 12
==> x2 = -11 ( Not valid because the number of discs can not be negative.
Then, there are 12 yellow discs in the bag.
Let the number of disks that are y be N. The number of disks that are b is given by 45 - N.
Now, two disks are removed from the bag with no replacement. The probability that the first is y is N/45. The probability that the second is y is (N - 1)/44
Therefore the probability that both are y is N(N-1)/44*45.
This is equal to 1/15.
Therefore N*(N-1)/ 44*45 = 1/15
=> N(N-1) = 44*45/15
=> N(N-1) = 44*3
=> N^2 - N - 132=0
=> N^2 - 12N + 11N - 132
=> N(N-12) + 11(N-12) = 0
=> (N + 11)(N -12) = 0
N can be 12 or -11, we ignore -11
Therefore the number of yellow disks in the bag is 12.
We can see that the probability of picking 2 yellow disks if there are 12 yellow disks in the 45 is given by (12/45)*(11/44) = 1/15.