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

Let C₁ and C₂ be two biased coins such that the probabilities of getting head in a single toss are 2/3 and 1/3, respectively. Suppose α is the number of heads that appear when C₁ is tossed twice, independently, and suppose β is the number of heads that appear when C₂ is tossed twice, independently. Then the probability that the roots of the quadratic polynomial x²-α x+β are real and equal, is

$P(H \text{ on } C_1) = \frac{2}{3},\qquad P(H \text{ on } C_2) = \frac{1}{3}$

$C_1 \text{ tossed twice} \to \alpha = \text{number of heads}$

$C_2 \text{ tossed twice} \to \beta = \text{number of heads}$

$\text{To find: } P(\text{roots of } x^2 - \alpha x + \beta \text{ are real and equal})$

$(1)\ \frac{40}{81} \qquad (2)\ \frac{20}{81} \qquad (3)\ \frac{1}{2} \qquad (4)\ \frac{1}{4}$

$1.\ \text{real and equal roots} \Leftrightarrow D = 0 \Leftrightarrow \alpha^2 = 4\beta$ $2.\ \text{distributions of } \alpha \sim B\!\left(2,\tfrac23\right),\ \beta \sim B\!\left(2,\tfrac13\right)$ $3.\ \text{valid } (\alpha,\beta),\ \text{then } P = \sum P(\alpha)\,P(\beta)$ $a=1,\quad b=-\alpha,\quad c=\beta$ $D = b^2 - 4ac = \alpha^2 - 4\beta$ $\text{equal roots} \Rightarrow D=0 \Rightarrow \alpha^2 = 4\beta$ $\alpha \in \{0,1,2\},\qquad \alpha \sim B\!\left(2,\ \tfrac23\right)$ $P(\alpha=0)=\left(\tfrac13\right)^2=\tfrac19$ $P(\alpha=1)=\binom{2}{1}\tfrac23\cdot\tfrac13=\tfrac49$ $P(\alpha=2)=\left(\tfrac23\right)^2=\tfrac49$ $\beta \in \{0,1,2\},\qquad \beta \sim B\!\left(2,\ \tfrac13\right)$ $P(\beta=0)=\left(\tfrac23\right)^2=\tfrac49$ $P(\beta=1)=\binom{2}{1}\tfrac13\cdot\tfrac23=\tfrac49$ $P(\beta=2)=\left(\tfrac13\right)^2=\tfrac19$ $\alpha=0:\ 0=4\beta \Rightarrow \beta=0 \quad \text{(valid)}$ $\alpha=1:\ 1=4\beta \Rightarrow \beta=\tfrac14 \quad \text{(reject)}$ $\alpha=2:\ 4=4\beta \Rightarrow \beta=1 \quad \text{(valid)}$ $(\alpha,\beta) \in \{(0,0),\ (2,1)\}$ $P(\alpha,\beta) = P(\alpha)\cdot P(\beta)$ $P = P(0,0) + P(2,1)$ $P(0,0) = \dfrac19\cdot\dfrac49 = \dfrac{4}{81}$ $P(2,1) = \dfrac49\cdot\dfrac49 = \dfrac{16}{81}$ $P = \dfrac{4}{81} + \dfrac{16}{81} = \dfrac{20}{81}$ $\boxed{\text{Answer: (2)}\quad \dfrac{20}{81}}$