Question

There are 10 marbles in a bag, and the marbles are either red or blue. Eric will randomly choose two marbles from the bag, without replacing the first one. If the probability of both marbles' being red is 2/15 , how many BLUE marbles are in the bag?
A) 3
B) 4
C) 5
D) 6

Answer 1

number of red marbles in bag = r.  b = 10-rProbability of first marble being red: r/10Second marble being red: (r-1)/9total probability = 2/15 = (r-1)*r/90r = 4b = 6

Answer 2

Answer:

The answer is option D = 6

Step-by-step explanation:

Let the number of red marbles be = r

Let the number of blue marbles be = 10-r

But as information about red marbles is given, so solving with respect to red (r)

$\frac{r}{10}*\frac{r-1}{9} =\frac{2}{15}$

Solving this we get:

$r^{2}-r-12=0$

$r^{2}-4r+3r-12=0$

$r(r-4)+3(r-4)=0$

This gives (r-4)=0 and (r+3)=0

r cannot be negative, so r = 4

Red marbles are = 4

Blue marbles are = 10-4 = 6

Hence, there are 6 blue marbles in the bag.