Let C_1 and C_2 be two biased coins such that | vedclass

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(B) For $C_1$,$P(H) = \frac{2}{3}$. The number of heads $\alpha$ follows a binomial distribution $B(2, \frac{2}{3})$.$P(\alpha = 0) = (\frac{1}{3})^2

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$\frac{20}{81}$

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(B) For $C_1$,$P(H) = \frac{2}{3}$. The number of heads $\alpha$ follows a binomial distribution $B(2, \frac{2}{3})$.$P(\alpha = 0) = (\frac{1}{3})^2 = \frac{1}{9}$$P(\alpha = 1) = 2 	imes \frac{2}{3} 	imes \frac{1}{3} = \frac{4}{9}$$P(\alpha = 2) = (\frac{2}{3})^2 = \frac{4}{9}$For $C_2$,$P(H) = \frac{1}{3}$. The number of heads $\beta$ follows a binomial distribution $B(2, \frac{1}{3})$.$P(\beta = 0) = (\frac{2}{3})^2 = \frac{4}{9}$$P(\beta = 1) = 2 	imes \frac{1}{3} 	imes \frac{2}{3} = \frac{4}{9}$$P(\beta = 2) = (\frac{1}{3})^2 = \frac{1}{9}$The roots of $x^2 - \alpha x + \beta = 0$ are real and equal if the discriminant $D = \alpha^2 - 4\beta = 0$,i.e.,$\alpha^2 = 4\beta$.Possible pairs $(\alpha, \beta)$ satisfying this are $(0, 0)$ and $(2, 1)$.Probability $= P(\alpha=0)P(\beta=0) + P(\alpha=2)P(\beta=1)$$= (\frac{1}{9} 	imes \frac{4}{9}) + (\frac{4}{9} 	imes \frac{4}{9}) = \frac{4}{81} + \frac{16}{81} = \frac{20}{81}$.

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