JEE Challenger
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MathMedAdv 2020 • Paper 1

Probability of Equal Roots in Quadratic Equation from Coin Flips

Consider two biased coins C1C_1 and C2C_2 with head probabilities 23\frac{2}{3} and 13\frac{1}{3} in a single flip, respectively. Let α\alpha denote the number of heads obtained when C1C_1 is flipped twice independently, and β\beta denote the number of heads obtained when C2C_2 is flipped twice independently. The probability that the quadratic polynomial x2−αx+βx^2 - \alpha x + \beta has real and equal roots is

Options

A

4081\frac{40}{81}

B

2081\frac{20}{81}

Correct
C

12\frac{1}{2}

D

14\frac{1}{4}

Topics & Concepts

Probability (Advanced)ProbabilityQuadratic Equations

Step-by-Step Solution

To find the probability that the quadratic polynomial x2−αx+βx^2 - \alpha x + \beta has real and equal roots, we analyze the conditions on the discriminant and the distributions of the random variables α\alpha and β\beta.

Step 1: Condition for Real and Equal Roots

A quadratic polynomial x2−αx+β=0x^2 - \alpha x + \beta = 0 has real and equal roots if and only if its discriminant is zero: Δ=(−α)2−4(1)(β)=0  ⟹  α2=4β\Delta = (-\alpha)^2 - 4(1)(\beta) = 0 \implies \alpha^2 = 4\beta

Step 2: Probability Distribution of α\alpha

The coin C1C_1 has a probability of getting heads p1=23p_1 = \frac{2}{3} and tails q1=13q_1 = \frac{1}{3}. When C1C_1 is flipped twice independently, the number of heads α∈{0,1,2}\alpha \in \{0, 1, 2\} follows a binomial distribution:

  • P(α=0)=(20)(13)2=19P(\alpha = 0) = \binom{2}{0} \left(\frac{1}{3}\right)^2 = \frac{1}{9}
  • P(α=1)=(21)(23)(13)=49P(\alpha = 1) = \binom{2}{1} \left(\frac{2}{3}\right)\left(\frac{1}{3}\right) = \frac{4}{9}
  • P(α=2)=(22)(23)2=49P(\alpha = 2) = \binom{2}{2} \left(\frac{2}{3}\right)^2 = \frac{4}{9}

Step 3: Probability Distribution of β\beta

The coin C2C_2 has a probability of getting heads p2=13p_2 = \frac{1}{3} and tails q2=23q_2 = \frac{2}{3}. When C2C_2 is flipped twice independently, the number of heads β∈{0,1,2}\beta \in \{0, 1, 2\} follows a binomial distribution:

  • P(β=0)=(20)(23)2=49P(\beta = 0) = \binom{2}{0} \left(\frac{2}{3}\right)^2 = \frac{4}{9}
  • P(β=1)=(21)(13)(23)=49P(\beta = 1) = \binom{2}{1} \left(\frac{1}{3}\right)\left(\frac{2}{3}\right) = \frac{4}{9}
  • P(β=2)=(22)(13)2=19P(\beta = 2) = \binom{2}{2} \left(\frac{1}{3}\right)^2 = \frac{1}{9}

Step 4: Finding Favorable Pairs (α,β)(\alpha, \beta)

We need to find the pairs (α,β)(\alpha, \beta) such that α2=4β\alpha^2 = 4\beta where α,β∈{0,1,2}\alpha, \beta \in \{0, 1, 2\}:

  1. For α=0\alpha = 0: 02=4β  ⟹  β=0  ⟹  (α,β)=(0,0)0^2 = 4\beta \implies \beta = 0 \implies (\alpha, \beta) = (0, 0)
  2. For α=1\alpha = 1: 12=4β  ⟹  β=14(Not possible since β∈{0,1,2})1^2 = 4\beta \implies \beta = \frac{1}{4} \quad (\text{Not possible since } \beta \in \{0, 1, 2\})
  3. For α=2\alpha = 2: 22=4β  ⟹  4=4β  ⟹  β=1  ⟹  (α,β)=(2,1)2^2 = 4\beta \implies 4 = 4\beta \implies \beta = 1 \implies (\alpha, \beta) = (2, 1)

Thus, the favorable pairs are (0,0)(0, 0) and (2,1)(2, 1).

Step 5: Calculating Total Probability

Since the flips of C1C_1 and C2C_2 are independent events, the total probability is: P(α2=4β)=P(α=0,β=0)+P(α=2,β=1)P(\alpha^2 = 4\beta) = P(\alpha = 0, \beta = 0) + P(\alpha = 2, \beta = 1)

P(α=0,β=0)=P(α=0)×P(β=0)=19×49=481P(\alpha = 0, \beta = 0) = P(\alpha = 0) \times P(\beta = 0) = \frac{1}{9} \times \frac{4}{9} = \frac{4}{81}

P(α=2,β=1)=P(α=2)×P(β=1)=49×49=1681P(\alpha = 2, \beta = 1) = P(\alpha = 2) \times P(\beta = 1) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}

Summing these probabilities: P(roots are real and equal)=481+1681=2081P(\text{roots are real and equal}) = \frac{4}{81} + \frac{16}{81} = \frac{20}{81}

Thus, the correct option is (B).

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