Question

Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all
previous coin tosses. Let 𝐸1 be the event that among the second, fourth, and sixth
coin tosses, there are at least two heads. Let 𝐸2 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(𝐸1| 𝐸2) is equal to ____________. (rounded off to
one decimal place)

Hint:

Toss 2 is shared between the events {2,4,6} and {1,2,3,5}. Condition on toss 2 being Head or Tail; the remaining tosses split into two independent groups, so compute each case separately and combine using \(P(E_1|E_2) = P(E_1 \cap E_2)/P(E_2)\).

Answer 1

Correct Answer: 0.5

We are given six independent fair coin tosses. Event \(E_1\): among tosses 2, 4, 6 there are at least two heads. Event \(E_2\): among tosses 1, 2, 3, 5 there are equal heads and tails, i.e. exactly two heads out of those four tosses. We need \(P(E_1 \mid E_2) = \dfrac{P(E_1 \cap E_2)}{P(E_2)}\).

Step 1: Identify the overlapping toss. \(E_1\) involves tosses {2, 4, 6} and \(E_2\) involves tosses {1, 2, 3, 5}. The only common toss is toss 2. Condition on the outcome of toss 2; the remaining tosses then split into two independent groups: {1, 3, 5} which determines \(E_2\), and {4, 6} which determines the rest of \(E_1\).

Step 2: Case toss 2 = Head, probability 1/2. For \(E_2\) to hold we now need exactly 1 head among tosses {1, 3, 5}, since toss 2 already supplies 1 head and the total required is 2:\[P(\text{1 head in 3 tosses}) = \binom{3}{1}\left(\frac{1}{2}\right)^3 = \frac{3}{8}\]For \(E_1\) we now need at least 1 head among {4, 6}, since toss 2 already contributes 1 toward the required 2:\[P(\text{at least 1 head in 2 tosses}) = 1 - \left(\frac{1}{2}\right)^2 = \frac{3}{4}\]Since {1,3,5} and {4,6} are independent, the joint contribution is:\[\frac{1}{2}\times\frac{3}{8}\times\frac{3}{4} = \frac{9}{64}\]

Step 3: Case toss 2 = Tail, probability 1/2. For \(E_2\) we now need exactly 2 heads among {1, 3, 5}:\[P(\text{2 heads in 3 tosses}) = \binom{3}{2}\left(\frac{1}{2}\right)^3 = \frac{3}{8}\]For \(E_1\) both toss 4 and toss 6 must be heads, since toss 2 contributes 0 heads and both required heads must come from {4, 6}:\[P(\text{2 heads in 2 tosses}) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\]Joint contribution:\[\frac{1}{2}\times\frac{3}{8}\times\frac{1}{4} = \frac{3}{64}\]

Step 4: Combine to get \(P(E_1 \cap E_2)\) and \(P(E_2)\).\[P(E_1 \cap E_2) = \frac{9}{64} + \frac{3}{64} = \frac{12}{64} = \frac{3}{16}\]\[P(E_2) = \frac{1}{2}\times\frac{3}{8} + \frac{1}{2}\times\frac{3}{8} = \frac{3}{8}\]

Step 5: Compute the conditional probability.\[P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{3/16}{3/8} = \frac{3}{16}\times\frac{8}{3} = \frac{1}{2} = 0.5\]

The final answer is: \[\boxed{0.5}\]