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?

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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.

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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.

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