Question

Let ${C_1}$ and ${C_2}$ be two biased coins such that the probabilities of getting head in a single toss are $\dfrac{2}{3}$and $\dfrac{1}{3}$, 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. $\dfrac{{40}}{{81}}$
B. $\dfrac{{20}}{{81}}$
C. $\dfrac{1}{2}$
D. $\dfrac{1}{4}$

Answer 1

Hint: In this question, we need to find the probability of the roots of quadratic equation ${x^2} + \alpha x + \beta $such that the roots are real and equal. First, we will find the probabilities of occurrence of tails in each biased coin. We know that, in a quadratic equation, the roots are real and equal when the determinant is zero. Now, in order to find the required probability, we will use permutation and combination concepts.

Formula Used:
The following formula will be useful for solving this question
$P(T) = 1 - P(H)$
$P(E) = {}^n{C_r}{P_r}$
where $RE,KE$are required event and known event respectively and ${}^n{C_r},P$ are combination and probability respectively.

Complete step by step solution:
 We know that for biased coin ${C_1}$,
$P{(H)_{{C_1}}} = \dfrac{2}{3}$
Therefore, by using formula $P(T) = 1 - P(H)$,we get
$ P{(T)_{{C_1}}} = 1 - \dfrac{2}{3} \\ \Rightarrow \dfrac{1}{3} $
Similarly, for biased coin ${C_2}$,$P{(H)_{{C_2}}} = \dfrac{1}{3}$,
Therefore, by using formula $P(T) = 1 - P(H)$,we get
$ P{(T)_{{C_2}}} = 1 - \dfrac{1}{3} \\ \Rightarrow \dfrac{2}{3} $
Now, we know that, in quadratic equations, the roots are real and equal only when the determinant of the equation is zero.
Therefore, for equation${x^2} + \alpha x + \beta $
$ D = 0 \\ {b^2} - 4ac = 0 \\ {\alpha ^2} - 4\beta = 0 \\ {\alpha ^2} = 4\beta $
which means that for $\alpha = 0$,$\beta $will be $0$and for $\alpha = 2$, $\beta $will be $1$
Therefore, by using formula $P(E) = {}^n{C_0} \times P_1^2 \times {}^n{C_0} \times P_2^2 + {}^n{C_n} \times P_1^2 \times {}^n{C_{n - 1}} \times {P_1} \times {P_2} + ...$, we get
$ P(E) = {}^n{C_0} \times P(T)_{{C_1}}^2 \times {}^n{C_0} \times P(T)_{{C_2}}^2 + {}^n{C_n} \times P(H)_{{C_1}}^2 \times {}^n{C_{n - 1}} \times P{(H)_{{C_1}}} \times P{(H)_{{C_2}}} \\ \Rightarrow {}^2{C_0} \times {\left( {\dfrac{1}{3}} \right)^2} \times {}^2{C_0} \times {\left( {\dfrac{2}{3}} \right)^2} + {}^2{C_2} \times {\left( {\dfrac{2}{3}} \right)^2} \times {}^2{C_1} \times \left( {\dfrac{1}{3}} \right) \times \left( {\dfrac{2}{3}} \right) \\ \Rightarrow \left( {\dfrac{1}{9}} \right) \times \left( {\dfrac{4}{9}} \right) + \left( {\dfrac{4}{9}} \right) \times \left( {\dfrac{4}{9}} \right) \\ \Rightarrow \dfrac{{20}}{{81}} $

Option ‘B’ is correct

Note: Students may make mistakes in selecting the values of events that occurred. Students should use a proper formula for probability which would be in terms of combination. This formula would be given in their textbooks or they can refer to the internet as well for the same.