A basket contains 3 gold-plated marbles, 4 diamond marbles, and some silver marbles, all of the same size and shape. Two marbles were drawn from the basket at random one after the other without replacement. If the probability that the two marbles were all silver is \(\frac{1}{15}\), find the number of silver marbles.
Explanation
Let the total number of marbles be x
No of gold-plated marbles, = 3, No of diamond marbles, = 4
No of silver marble = x - 3 - 4 = (x - 7)
Pr(both were silver) = \(\frac{(x - 7)}{x}\) x \(\frac{(x - 8)}{(x - 1)}\) ( since selection is without replacement will affect the second pick)
= \(\frac{(x - 7)}{x}\) x \(\frac{(x - 8)}{(x - 1)}\) = \(\frac{1}{15}\)
\(\frac{(x^2 - 15x + 56 )}{(x^2 - x) }\) = \(\frac{1}{15}\),
cross multiply
= 15x\(^2\) - 225x + 840 = x\(^2\) - x
= 14x\(^2\) - 224x + 840 = 0
Divide through by 14
= x\(^2\) - 16x + 60 = 0
= x\(^2\) - 10x - 6x + 60 = 0
= x(x - 10) - 6(x - 10) = 0
= (x - 10)(x - 6) = 0
= x = 6 or 10
x = 10 ( total marble should not be less than the sum of both gold and diamond marbles)
Recall, Silver marbles = x - 7 = 10 - 7 = 3
THEREFORE, the number of silver marbles = 3.