Question

If a coin is tossed seven times, then the probability of getting exactly three heads such that no two heads occur consecutively is

• A. $\frac{5}{64}$
• B. $\frac{5}{32}$
• C. $\frac{5}{128}$
• D. $\frac{35}{128}$

Hint:

The "gaps" method is the go-to technique for problems involving non-consecutive arrangements. To arrange $k$ items non-consecutively among $m$ other items, first place the $m$ items, creating $m+1$ gaps. Then, choose $k$ of these gaps to place the restricted items. The number of ways is $^{m+1}C_k$.

Answer 1

The Correct Option is B

Step 1: Total Outcomes:
Tossing a coin 7 times gives 27 = 128 total possible outcomes.

Step 2: Favorable Outcomes (Gap Method):
We need exactly 3 Heads (H) and 4 Tails (T) such that no two Heads are consecutive.

First, place the 4 Tails:

T T   T   T

This creates 5 possible spaces (gaps) where Heads can be placed (marked by _):

_ T _ T _ T _ T _

To ensure no two Heads are together, we must place the 3 Heads in 3 distinct gaps out of the 5 available.

Number of ways = C(5,3)

C(5,3) = (5 × 4 × 3) / (3 × 2 × 1) = 10

Step 3: Calculate Probability:

P(E) = Favorable / Total = 10 / 128 = 5 / 64