Correct : 0.5

The correct answer is 0.5.
This is a conditional probability problem involving 6 independent fair coin tosses (T1 through T6). We need to find P(E₁ | E₂), where E₁ and E₂ are defined over overlapping but not identical subsets of tosses.
E₁ - among tosses 2, 4, 6, there are at least 2 heads. E₂ - among tosses 1, 2, 3, 5, there are equal heads and tails (exactly 2 heads, 2 tails).
Key Insight - Shared toss T2: Toss 2 is the only toss common to both events. Tosses T4 and T6 are entirely outside E₂'s scope, and tosses T1, T3, T5 are entirely outside E₁'s scope. Since all tosses are independent, T4 and T6 remain unconditionally fair (P = 1/2 each) even given E₂.
Conditional distribution of T2 given E₂: By symmetry of the 4 tosses in E₂, each individual toss is equally likely to be a head. So P(T2 = H | E₂) = 2/4 = 1/2 - the same as the unconditional probability.
Computing P(E₁ | E₂): Given E₂, the three tosses {T2, T4, T6} each independently show heads with probability 1/2. Therefore, the number of heads in {T2, T4, T6} given E₂ follows a Binomial(3, 1/2) distribution - identical to the unconditional case.
P(E₁ | E₂) = P(at least 2 heads in 3 fair tosses) = P(X=2) + P(X=3) = C(3,2)·(1/2)³ + C(3,3)·(1/2)³ = 3/8 + 1/8 = 4/8 = 0.5.