# One bag contains 2 red marbles and 2 blue marbles. A second bag contains 2 redmarbles, 2 blue marbles, and g green marbles, with g > 0. For each bag, Mariacalculates the probability of randomly drawing two marbles of the same colour in two draws from that bag, without replacement. (Drawing two marbles without replacement means drawing two marbles, one after the other, without putting the first marble back into the bag.) If these two probabilities are equal, then the value of g is (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

1
by Deleted account

2016-04-06T22:32:48+05:30
For First bag,

Total marbles = 2 + 2 = 4

We are taking out two marbles one after another without replacement,

So, total possible ways of getting two blue marbles = 2 x 1 = 2
Total possible ways of getting two red marbles = 2 x 1 = 2

Possibility of getting same colour marbles = 2 + 2 = 4

But, possible ways of getting any two marbles = 4 x (4-1) = 4 x 3 = 12

So, probability of getting same colour marbles both times = 4/12 = 1/3

For second bag,

Total marbles = 2 + 2 + g = 4 + g

We are taking out two marbles one after another without replacement,

So, total possible ways of getting two blue marbles = 2 x 1 = 2
Total possible ways of getting two red marbles = 2 x 1 = 2
Total possible ways of getting two red marbles = g x (g-1)

Possibility of getting same colour marbles = 2 + 2 + g(g-1) = 4 + g
² - g

But, possible ways of getting any two marbles = g+4 x [(g-1)+4] = (g+4)(g+3)

So, probability of getting same colour marbles both times = 4 +g
² - g/ (g+4)(g+3)

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Now, given that the probabilities are same for both the bags,

1/3 = 4 +g² -g/(g+4)(g+3)
1/12+3g² - 3g = 1/(g+4)(g+3)
12+3g² - 3g = (g+4)(g+3)
12+3g² - 3g = g² +7g + 12
2g² = 10g
0 = 2g² - 10g
0 = 2g(g-5)

By factor theorem,
g = 0
g = 5

Since, it's given that g > 0, g = 5 in this case.

So, the answer is (B) 5.