A bag contains only red and blue marbles. Yasmine takes one marble at random from the bag. The probability that she takes a red marble is 1 in 5. Yasmine returns the marble to the bag and adds five more red marbles to the bag. The probability that she takes one red marble at random is now 1 in 3. How many red marbles were originally in the bag?
A. 3 red
B. 5 red
C. 10 red
D. 2 red
Oops i forgot to put i though B was the right answer ( B. 5 red
thought ^^
r/(r+b) = 1/5
(r+5)/(r+5+b) = 1/3
r=5
you are correct
To solve this problem, let's work through it step by step and use algebra to find the answer. Let's start by representing the number of red marbles in the original bag as "R."
According to the problem, the probability that Yasmine takes a red marble from the original bag is 1 in 5. This means that the ratio of the number of red marbles to the total number of marbles in the original bag is 1/5, or R/(R + B) = 1/5, where "B" represents the number of blue marbles.
Next, Yasmine returns the marble to the bag and adds five more red marbles. The new total number of red marbles becomes (R + 5), and the total number of marbles becomes (R + B + 5). The probability that Yasmine takes a red marble from this new bag is now 1 in 3, which means the ratio of red marbles to total marbles is (R + 5)/(R + B + 5) = 1/3.
Now we have two equations:
1) R/(R + B) = 1/5
2) (R + 5)/(R + B + 5) = 1/3
We can simplify these equations by cross-multiplying:
1) 5R = R + B
2) 3(R + 5) = R + B + 5
Let's solve these equations:
1) 5R = R + B
4R = B (subtracting R from both sides)
2) 3(R + 5) = R + B + 5
3R + 15 = R + B + 5
3R - R = B - 10 (subtracting B + 5 from both sides)
2R = B - 10
Now we have a system of equations:
4R = B (Equation 1)
2R = B - 10 (Equation 2)
We can substitute Equation 1 into Equation 2 to solve for R:
2R = 4R - 10
10 = 2R (subtracting 4R from both sides)
R = 5
Therefore, the number of red marbles originally in the bag is 5. So the correct answer is B.