Question

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

• A. $\frac{40}{81}$
• B. $\frac{20}{81}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B

$\frac{20}{81}$

Explanation for the correct option:

Determining the probability that the roots of the quadratic polynomial $x^2-\alpha x+\beta$ are real and equal:

Consider the roots of the equation $x^2-\alpha x+\beta$ are real and equal

When, $D=0$

$$\begin{gathered} {\alpha }^2-4\beta =0 \\ {\alpha }^2=4\beta \end{gathered}$$

Case 1: $\left(\alpha =0,\beta =0\right)$

We know that

$p\left(x=r\right)=\sum _{r=0}^n{}^{n}C_r{\left(p\right)}^r{\left(q\right)}^{n-r}$

$$\begin{gathered} P_1={}^{2}C_0{\left(\frac{2}{3}\right)}^0{\left(\frac{1}{3}\right)}^2\times {}^{2}C_0{\left(\frac{1}{3}\right)}^0{\left(\frac{2}{3}\right)}^2 \\ {P}_1=\frac{4}{81} \end{gathered}$$

Case 2: $\left(\alpha =2,\beta =1\right)$

$$\begin{gathered} P_2={}^{2}C_2{\left(\frac{2}{3}\right)}^2\times {}^{2}C_1{\left(\frac{1}{3}\right)}^1{\left(\frac{2}{3}\right)}^1 \\ {P}_2=\frac{16}{81} \end{gathered}$$

Thus, the required probability is:

$$\begin{gathered} P_1+P_2=\frac{4}{81}+\frac{16}{81} \\ {P}_1+P_2=\frac{20}{81} \end{gathered}$$

Hence, option (B) is the correct answer.