Triscia had 12 jellybeans in a jar (orange ones, green ones, and purple ones). After she ate all the orange ones (but none of any other color), the probability of taking out a green jellybean by chance was twice as high as it was before she ate any jellybeans. How many orange jellybeans were there to start with, if 4 of her jellybeans were purple?
12/2=6 6/2=3
So the answer is 6
To solve this problem, we can start by setting up a system of equations based on the given information.
Let's assume that Triscia initially had "o" orange jellybeans, "g" green jellybeans, and 4 purple jellybeans. Since Triscia ate all the orange ones, the number of green and purple jellybeans remains the same.
Given that the probability of selecting a green jellybean after eating all the orange ones is twice as high as before, we can set up the following equation:
(g/8) = 2(g/(o + g + 4))
The left side represents the probability of selecting a green jellybean after eating all the orange ones. The right side represents the probability of selecting a green jellybean before eating any jellybeans (which is g/(o + g + 4)), multiplied by 2 to reflect the increased probability.
Now, let's solve for o:
(g/8) = 2g/(o + g + 4)
Multiplying both sides by (o + g + 4) to remove the denominator:
g*(o + g + 4) = 2g*8
o*g + g^2 + 4g = 16g
Rearranging the equation:
o*g + g^2 -12g = 0
Factoring out "g":
g*(o + g - 12) = 0
Since we know that the total number of jellybeans must be positive, g cannot be zero. Therefore, we have:
o + g - 12 = 0
Substituting the value of g, we have:
o + 12 - 12 = 0
o = 0
Therefore, there were 0 orange jellybeans to start with.
It seems there might be a mistake or missing information in the problem statement, as we end up with 0 orange jellybeans.