Let C1 and C2 be two biased coins such that the probabilities of getting head in a single toss are $\frac{3}{4}$ and $\frac{1}{4}$ respectively. Suppose $α$ is the number of heads that appear when $C_{1}$ is tossed twice, independently, and suppose $β$ is the number of heads that appear when $C_{2}$ is tossed twice, independently, Then probability that the roots of the quadratic polynomial $x^{2} - α x + β$ are real and equal, is

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$\frac{57}{256}$

$\frac{63}{256}$

$\frac{15}{256}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

$x^{2} - α x + β$
Roots are real and equal $\Rightarrow α^{2} = 4 β$
i) $β = 0 \Rightarrow α = 0 \Rightarrow (0 H, 0 H) \Rightarrow [^{2} C_{0} {(\frac{3}{4})}^{0} {(\frac{1}{4})}^{2}] [^{2} C_{0} {(\frac{1}{4})}^{0} {(\frac{3}{4})}^{2}] = \frac{1^{2} {.3}^{2}}{4^{4}}$
ii) $β = 1 \Rightarrow α = 2 \Rightarrow (2 H, 1 H) \Rightarrow [^{2} C_{2} {(\frac{3}{4})}^{2} {(\frac{1}{4})}^{0}] [^{2} C_{1} {(\frac{1}{4})}^{1} {(\frac{3}{4})}^{1}] = \frac{3^{2} .2.1.3}{4^{4}}$
Required probability = $\frac{9}{256} + \frac{54}{256} = \frac{63}{256}$