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 $P(E_1 \mid E_2)$ is equal to ____________. (rounded off to one decimal place)

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Step 1: Set up the sample space and define the events. An unbiased coin is tossed 6 times, and each toss is independent, so every one of the $$ 2^6 = 64 $$ outcomes is equally likely. Let $X_1, X_2, X_3, X_4, X_5, X_6$ denote the outcome of the first through sixth tosses, where each $X_i$ is Heads or Tails with probability $\frac{1}{2}$ each. Event $E_1$: among the second, fourth, and sixth tosses ($X_2, X_4, X_6$), there are at least two heads. Event $E_2$: among the first, second, third, and fifth tosses ($X_1, X_2, X_3, X_5$), there are an equal number of heads and tails, that is, exactly $2$ heads and $2$ tails among these four tosses. Step 2: Notice the shared toss between the two events. $E_1$ depends on $X_2, X_4, X_6$, and $E_2$ depends on $X_1, X_2, X_3, X_5$. The only toss common to both events is $X_2$. The tosses $X_4, X_6$ used in $E_1$ do not appear in $E_2$ at all, and the tosses $X_1, X_3, X_5$ used in $E_2$ do not appear in $E_1$ at all. Because all six tosses are independent, we can break both $P(E_2)$ and $P(E_1 \cap E_2)$ into pieces by conditioning on the outcome of $X_2$. Step 3: Compute $P(E_2)$ directly. $E_2$ needs exactly $2$ heads out of the $4$ tosses $X_1, X_2, X_3, X_5$, which are all independent fair coins. $$ P(E_2) = \binom{4}{2}\left(\frac{1}{2}\right)^4 = \frac{6}{16} = \frac{3}{8} $$ Step 4: Split $P(E_2)$ by the value of $X_2$, as a check. If $X_2 = \text{Head}$, then $E_2$ needs exactly $1$ head among the remaining three tosses $X_1, X_3, X_5$: $$ P(\text{exactly 1 head in } X_1, X_3, X_5) = \binom{3}{1}\left(\frac{1}{2}\right)^3 = \frac{3}{8} $$ If $X_2 = \text{Tail}$, then $E_2$ needs exactly $2$ heads among $X_1, X_3, X_5$: $$ P(\text{exactly 2 heads in } X_1, X_3, X_5) = \binom{3}{2}\left(\frac{1}{2}\right)^3 = \frac{3}{8} $$ Since $P(X_2 = \text{Head}) = P(X_2 = \text{Tail}) = \frac{1}{2}$, $$ P(E_2) = \frac{1}{2}\cdot\frac{3}{8} + \frac{1}{2}\cdot\frac{3}{8} = \frac{3}{8} $$ which matches Step 3, confirming the setup is right. Step 5: Work out what $E_1$ needs, depending on $X_2$. $E_1$ needs at least $2$ heads among $X_2, X_4, X_6$ (three tosses). If $X_2 = \text{Head}$, one head is already secured from $X_2$, so $E_1$ needs at least $1$ more head from $X_4, X_6$ together, that is, at least one of $X_4, X_6$ is heads. $$ P(\text{at least 1 head in } X_4, X_6) = 1 - P(\text{no heads in } X_4, X_6) = 1 - \frac{1}{4} = \frac{3}{4} $$ If $X_2 = \text{Tail}$, zero heads are secured from $X_2$, so $E_1$ needs both $X_4$ and $X_6$ to be heads: $$ P(X_4 = \text{Head and } X_6 = \text{Head}) = \frac{1}{4} $$ Step 6: Compute $P(E_1 \cap E_2)$ by combining the pieces for each value of $X_2$. Because $X_4, X_6$ are independent of $X_1, X_3, X_5$, the joint probability for each branch of $X_2$ is the product of the piece from Step 4 and the piece from Step 5. For $X_2 = \text{Head}$: $$ \frac{1}{2}\cdot\frac{3}{8}\cdot\frac{3}{4} = \frac{9}{64} $$ For $X_2 = \text{Tail}$: $$ \frac{1}{2}\cdot\frac{3}{8}\cdot\frac{1}{4} = \frac{3}{64} $$ Adding the two branches, $$ P(E_1 \cap E_2) = \frac{9}{64} + \frac{3}{64} = \frac{12}{64} = \frac{3}{16} $$ Step 7: Compute the conditional probability. $$ P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{3/16}{3/8} = \frac{3}{16}\times\frac{8}{3} = \frac{8}{16} = \frac{1}{2} $$ Step 8: Final Answer. Rounded off to one decimal place, the conditional probability is 0.5. $$ \boxed{0.5} $$

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Correct Answer: 0.5

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