Let `C_(1) and C_(2)` be two biased coins such that the probabilities of getting heat in a single toss are `(2)/(3) and (1)/(3)`, respectively. Suppose `alpha` is the number of hed that appear when `C_(1)` is tossed twice. Independently. Then the probability that the roots of the quadratic polynomil `x^(2)-ax+beta` are real equals, is

Options

A. `(40)/(81)`
B. `(20)/(81)`
C. `(1)/(2)`
D. `(1)/(4)`

Answer

To solve the problem, we need to find the probability that the roots of the quadratic polynomial \( x^2 - \alpha x + \beta \) are real and equal. This occurs when the discriminant \( D \) is equal to zero, which leads us to the condition:

\[ D = \alpha^2 - 4\beta = 0 \implies \alpha^2 = 4\beta \]

### Step 1: Determine possible values of \( \alpha \) and \( \beta \)

1. **Calculate \( \alpha \)**:
- \( \alpha \) is the number of heads when coin \( C_1 \) (with probability of heads \( \frac{2}{3} \)) is tossed twice.
- Possible values of \( \alpha \) are \( 0, 1, 2 \).

2. **Calculate \( \beta \)**:
- \( \beta \) is the number of heads when coin \( C_2 \) (with probability of heads \( \frac{1}{3} \)) is tossed twice.
- Possible values of \( \beta \) are \( 0, 1, 2 \).

### Step 2: Find pairs \( (\alpha, \beta) \) such that \( \alpha^2 = 4\beta \)

We will check each possible value of \( \alpha \):

- **Case 1**: \( \alpha = 0 \)
\[ 0^2 = 4\beta \implies \beta = 0 \]
Valid pair: \( (0, 0) \)

- **Case 2**: \( \alpha = 1 \)
\[ 1^2 = 4\beta \implies \beta = \frac{1}{4} \quad \text{(not valid since } \beta \text{ must be an integer)} \]

- **Case 3**: \( \alpha = 2 \)
\[ 2^2 = 4\beta \implies \beta = 1 \]
Valid pair: \( (2, 1) \)

### Step 3: Calculate probabilities for valid pairs

1. **For \( (\alpha, \beta) = (0, 0) \)**:
- Probability of getting \( \alpha = 0 \) (no heads in 2 tosses of \( C_1 \)):
\[ P(\alpha = 0) = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \]
- Probability of getting \( \beta = 0 \) (no heads in 2 tosses of \( C_2 \)):
\[ P(\beta = 0) = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]
- Combined probability:
\[ P(0, 0) = P(\alpha = 0) \cdot P(\beta = 0) = \frac{1}{9} \cdot \frac{4}{9} = \frac{4}{81} \]

2. **For \( (\alpha, \beta) = (2, 1) \)**:
- Probability of getting \( \alpha = 2 \) (two heads in 2 tosses of \( C_1 \)):
\[ P(\alpha = 2) = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]
- Probability of getting \( \beta = 1 \) (one head in 2 tosses of \( C_2 \)):
\[ P(\beta = 1) = 2 \cdot \left( \frac{1}{3} \right)^1 \cdot \left( \frac{2}{3} \right)^1 = 2 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9} \]
- Combined probability:
\[ P(2, 1) = P(\alpha = 2) \cdot P(\beta = 1) = \frac{4}{9} \cdot \frac{4}{9} = \frac{16}{81} \]

### Step 4: Total probability

The total probability that the roots of the polynomial are real and equal is given by the sum of the probabilities of the valid pairs:

\[ P(\text{real and equal roots}) = P(0, 0) + P(2, 1) = \frac{4}{81} + \frac{16}{81} = \frac{20}{81} \]

### Final Answer

The probability that the roots of the quadratic polynomial \( x^2 - \alpha x + \beta \) are real and equal is:

\[ \boxed{\frac{20}{81}} \]