Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all previous coin tosses. Let $E_1$ be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let $E_2$ be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails. The conditional probability $P(E_1|E_2)$ is equal to __________. (rounded off to one decimal place)

Let the outcomes of the 6 tosses be denoted by $T_1, T_2, T_3, T_4, T_5, T_6$.

• $E_1$: Event that among $\{T_2, T_4, T_6\}$, there are at least 2 heads.

• $E_2$: Event that among $\{T_1, T_2, T_3, T_5\}$, there are exactly 2 heads and 2 tails.

Notice that only the toss $T_2$ is shared between $E_1$ and $E_2$.

We condition on the outcome of $T_2$:

Case 1: $T_2 = H$ (probability $= 1/2$)

• For $E_2$ to occur: Among the remaining 3 tosses $\{T_1, T_3, T_5\}$, exactly 1 head must appear.

  $$P(E_2 \mid T_2 = H) = \frac{\binom{3}{1}}{2^3} = \frac{3}{8}$$

• For $E_1$ to occur: Among $\{T_4, T_6\}$, at least 1 head must appear.

  $$P(E_1 \mid T_2 = H) = 1 - P(\text{0 heads}) = 1 - \frac{1}{4} = \frac{3}{4}$$

• Since $\{T_4, T_6\}$ are independent of $\{T_1, T_3, T_5\}$:

  $$P(E_1 \cap E_2 \mid T_2 = H) = \frac{3}{4} \times \frac{3}{8} = \frac{9}{32}$$

Case 2: $T_2 = T$ (probability $= 1/2$)

• For $E_2$ to occur: Among $\{T_1, T_3, T_5\}$, exactly 2 heads must appear.

  $$P(E_2 \mid T_2 = T) = \frac{\binom{3}{2}}{2^3} = \frac{3}{8}$$

• For $E_1$ to occur: Among $\{T_4, T_6\}$, both tosses must be heads.

  $$P(E_1 \mid T_2 = T) = \frac{1}{4}$$

• Hence:

  $$P(E_1 \cap E_2 \mid T_2 = T) = \frac{1}{4} \times \frac{3}{8} = \frac{3}{32}$$

Computing the Probabilities:

Conditional Probability: