Question

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)

Hint:

Condition on the outcome of the second toss, since it is shared by both events, then combine the two branches using the independence of the remaining tosses.

Answer 1

Correct Answer: 0.5

Step 1: Count total equally likely outcomes.
Six fair coin tosses give $2^6=64$ equally likely outcomes in total. Instead of multiplying probabilities, this route counts outcomes directly, which gives a different way to reach the same answer.

Step 2: Restate what each event needs.
$E_1$ needs at least $2$ heads among tosses $2, 4, 6$. $E_2$ needs exactly $2$ heads and $2$ tails among tosses $1, 2, 3, 5$. Toss $2$ is shared by both events, while tosses $1, 3, 5$ belong only to $E_2$ and tosses $4, 6$ belong only to $E_1$.

Step 3: Count outcomes satisfying $E_2$.
Among the four tosses $1,2,3,5$, we need exactly $2$ heads out of $4$, which can happen in $\binom{4}{2}=6$ ways. Tosses $4$ and $6$ are free to be anything, giving $4$ combinations. So the number of outcomes satisfying $E_2$ is
\[ 6 \times 4 = 24 \]
out of $64$, giving $P(E_2)=24/64=3/8$, matching the direct binomial count.

Step 4: Split the $6$ arrangements of $E_2$ by whether toss $2$ is a head.
Out of the $\binom{4}{2}=6$ ways to place $2$ heads among tosses $1,2,3,5$: in $\binom{3}{1}=3$ of them toss $2$ is a head (with exactly $1$ of the remaining tosses $1,3,5$ also a head), and in $\binom{3}{2}=3$ of them toss $2$ is a tail (with exactly $2$ of $1,3,5$ heads).

Step 5: Count how tosses $4,6$ must fall for $E_1$ to hold, in each branch.
If toss $2$ is a head (the $3$ arrangements from Step 4), $E_1$ needs at least $1$ more head among tosses $4,6$. Out of the $4$ combinations of $(X_4,X_6)$, only $TT$ fails, so $3$ combinations work.
If toss $2$ is a tail (the other $3$ arrangements), $E_1$ needs both toss $4$ and toss $6$ to be heads, so only $1$ combination out of $4$ works.

Step 6: Multiply within each branch and add.
Head branch: $3$ arrangements of $1,2,3,5$ times $3$ working combinations of $4,6$ gives $9$ outcomes.
Tail branch: $3$ arrangements of $1,2,3,5$ times $1$ working combination of $4,6$ gives $3$ outcomes.
Total outcomes satisfying both $E_1$ and $E_2$:
\[ 9 + 3 = 12 \]

Step 7: Form the conditional probability as a ratio of counts.
\[ P(E_1 \mid E_2) = \frac{\text{outcomes in } E_1 \cap E_2}{\text{outcomes in } E_2} = \frac{12}{24} = \frac{1}{2} \]

Step 8: State the rounded value.
This is $0.5$ when rounded to one decimal place, matching the probability based calculation.
\[ \boxed{0.5} \]