JEE Advanced 2020 — Paper 1 — Question 33

For real and equal roots, discriminant $\alpha^2 - 4\beta = 0$, so $\alpha^2 = 4\beta$. Since $\alpha, \beta$ are non-negative integers from 0 to 2, the only possibilities are $(\alpha=0,\beta=0)$ and $(\alpha=2,\beta=1)$. For coin $C_1$, $P(H)=\frac{2}{3}$, $P(T)=\frac{1}{3}$. $P(\alpha=0) = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$. $P(\alpha=2) = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$. For coin $C_2$, $P(H)=\frac{1}{3}$, $P(T)=\frac{2}{3}$. $P(\beta=0) = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$. $P(\beta=1) = \binom{2}{1}\left(\frac{1}{3}\right)\left(\frac{2}{3}\right) = 2 \cdot \frac{2}{9} = \frac{4}{9}$. Since the tosses are independent, $P(\alpha=0,\beta=0) = \frac{1}{9} \cdot \frac{4}{9} = \frac{4}{81}$. $P(\alpha=2,\beta=1) = \frac{4}{9} \cdot \frac{4}{9} = \frac{16}{81}$. Total probability = $\frac{4}{81} + \frac{16}{81} = \frac{20}{81}$.