4. ten green marbles are added to a bag containing an unknown number of red marbles. suppose two marbles are…

Explanation:

Step1: Calculate the experimental - probability of choosing 2 red marbles

The experiment is repeated 9 times, and 2 red marbles are chosen 4 times. So the experimental - probability $P$ of choosing 2 red marbles is $P=\frac{4}{9}$.

Step2: Set up the probability formula for choosing 2 red marbles

Let the number of red marbles be $x$. The total number of marbles in the bag is $x + 10$. The probability of choosing 2 red marbles on the first draw is $\frac{x}{x + 10}$, and on the second draw (without replacement) is $\frac{x-1}{x + 9}$. So the probability of choosing 2 red marbles is $P=\frac{x(x - 1)}{(x + 10)(x + 9)}$.

Step3: Equate the experimental and theoretical probabilities

We have $\frac{x(x - 1)}{(x + 10)(x + 9)}=\frac{4}{9}$. Cross - multiply: $9x(x - 1)=4(x + 10)(x + 9)$. Expand both sides: $9x^{2}-9x=4(x^{2}+9x + 10x+90)$. $9x^{2}-9x=4x^{2}+36x + 40x+360$. $9x^{2}-9x=4x^{2}+76x + 360$. Move all terms to one side: $9x^{2}-4x^{2}-9x-76x - 360 = 0$. $5x^{2}-85x - 360 = 0$. Divide through by 5: $x^{2}-17x - 72 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b=-17$, $c=-72$), we use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-17)^{2}-4\times1\times(-72)=289 + 288 = 577$. $x=\frac{17\pm\sqrt{577}}{2}$. We can also factor the quadratic equation $x^{2}-17x - 72=(x - 24)(x + 3)=0$. Setting each factor equal to zero gives $x = 24$ or $x=-3$. Since the number of marbles cannot be negative, $x = 24$.

Answer:

24