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

Instead of conditioning step by step, count favourable equally likely sequences directly out of all \(2^6 = 64\) outcomes of the six tosses.

Step 1: Count sequences satisfying \(E_2\). \(E_2\) needs exactly 2 heads among the 4 positions {1, 2, 3, 5}. Number of ways to pick which 2 of these 4 positions are heads is \(\binom{4}{2} = 6\). Positions 4 and 6 are unrestricted by \(E_2\), giving \(2^2 = 4\) free choices. So the number of sequences with \(E_2\) true is \(6 \times 4 = 24\), hence \(P(E_2) = 24/64 = 3/8\).

Step 2: Count sequences satisfying both \(E_1\) and \(E_2\), split by the value at position 2. Position 2 is shared between the two events, so split on it.

If position 2 = Head: \(E_2\) then requires exactly 1 more head among {1, 3, 5}, giving \(\binom{3}{1} = 3\) arrangements. For \(E_1\), positions {2,4,6} already have 1 head from position 2, so we need at least 1 head among {4, 6}; out of 4 possible (toss4, toss6) pairs, only TT fails, so 3 pairs work. This case contributes \(3 \times 3 = 9\) sequences.

If position 2 = Tail: \(E_2\) then requires exactly 2 heads among {1, 3, 5}, giving \(\binom{3}{2} = 3\) arrangements. For \(E_1\), position 2 contributes 0 heads, so both toss4 and toss6 must be heads to reach at least 2 heads among {2,4,6}; only 1 such pair (HH) works. This case contributes \(3 \times 1 = 3\) sequences.

Step 3: Total favourable count. Sequences satisfying \(E_1 \cap E_2\) = \(9 + 3 = 12\) out of 64, so \(P(E_1 \cap E_2) = 12/64 = 3/16\).

Step 4: Apply the conditional probability formula.\[P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{12/64}{24/64} = \frac{12}{24} = \frac{1}{2}\]This confirms the answer as \(0.5\), matching the case-based approach exactly, since both methods count the same 12 out of 24 equally likely outcomes restricted to \(E_2\).

Final answer: \(0.5\)