Let $$C_{1}$$ and $$C_{2}$$ be two biased coins such that the probabilities of getting head in a single toss are $$\frac{2}{3}$$ and $$\frac{1}{3}$$ , respectively. Suppose $$\alpha$$ is the number of heads that appear when $$C_{1}$$ is tossed twice, independently, and suppose $$\beta$$ is the number of heads that appear when $$C_{2}$$ is tossed twice, independently. Then the probability that the roots of the quadratic polynomial $$x^{2} − \alpha x + \beta$$ are real and equal, is

For any quadratic $$x^{2}+bx+c$$, the roots are real and equal (a repeated root) exactly when the discriminant is zero:

$$b^{2}-4ac = 0$$

In this problem the quadratic is $$x^{2}-\alpha x+\beta$$, so

$$a=1,\; b=-\alpha,\; c=\beta$$

Setting the discriminant to zero gives

$$(-\alpha)^{2}-4(1)(\beta)=0$$ $$\Rightarrow\; \alpha^{2}=4\beta$$ $$\Rightarrow\; \beta=\dfrac{\alpha^{2}}{4}$$

Because each coin is tossed twice, both $$\alpha$$ and $$\beta$$ can take the integer values $$0,1,2$$ only. Hence we list the pairs $$(\alpha,\beta)$$ that satisfy $$\beta=\alpha^{2}/4$$ within this set:

• $$\alpha=0 \;\Rightarrow\; \beta=0$$ • $$\alpha=2 \;\Rightarrow\; \beta=1$$ (The case $$\alpha=1$$ leads to $$\beta=1/4$$, which is impossible.)

Next, compute the required probabilities. All tosses are independent.

Probability for $$\alpha$$ (coin $$C_1$$, head-probability $$\tfrac{2}{3}$$):
  • $$P(\alpha=0)=(1-\tfrac{2}{3})^{2}=(\tfrac{1}{3})^{2}=\tfrac{1}{9}$$
  • $$P(\alpha=2)=(\tfrac{2}{3})^{2}=\tfrac{4}{9}$$

Probability for $$\beta$$ (coin $$C_2$$, head-probability $$\tfrac{1}{3}$$):
  • $$P(\beta=0)=(1-\tfrac{1}{3})^{2}=(\tfrac{2}{3})^{2}=\tfrac{4}{9}$$
  • $$P(\beta=1)=2\cdot\tfrac{1}{3}\cdot\tfrac{2}{3}=\tfrac{4}{9}$$

Total probability for real and equal roots:

$$\begin{aligned} P &= P(\alpha=0,\beta=0)+P(\alpha=2,\beta=1)\\ &= P(\alpha=0)\,P(\beta=0)+P(\alpha=2)\,P(\beta=1)\\ &= \tfrac{1}{9}\cdot\tfrac{4}{9}+\tfrac{4}{9}\cdot\tfrac{4}{9}\\ &= \tfrac{4}{81}+\tfrac{16}{81}\\ &= \tfrac{20}{81} \end{aligned}$$

Therefore, the required probability is $$\tfrac{20}{81}$$.

Option B which is: $$\dfrac{20}{81}$$