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: Understanding the Concept
This is a probability problem combining concepts of binomial probability and combinatorial arrangements with restrictions. We need to find the number of specific arrangements (3 heads, 4 tails, no consecutive heads) and divide it by the total number of possible outcomes.
Step 2: Key Formula or Approach
1. Total Outcomes: For 7 coin tosses, the total number of possible outcomes is $2^7$. 2. Favorable Outcomes: We need to find the number of ways to arrange 3 Heads (H) and 4 Tails (T) such that no two H's are together. This is a classic "gaps" problem. 3. Probability = (Favorable Outcomes) / (Total Outcomes).
Step 3: Detailed Explanation
1. Total Outcomes: A fair coin is tossed 7 times. Each toss has 2 possible outcomes (H or T). Total number of outcomes = $2^7 = 128$. 2. Favorable Outcomes: We want to arrange 3 H's and 4 T's so that no two H's are consecutive. First, place the 4 Tails in a row. This creates gaps where the Heads can be placed. \[ \_ T \_ T \_ T \_ T \_ \] There are 4 T's, which create $4+1 = 5$ possible gaps (indicated by \_). To ensure that no two Heads are consecutive, we must place each of the 3 Heads in a different gap. The number of ways to choose 3 distinct gaps out of the 5 available gaps is given by the combination formula: \[ \text{Number of ways} = {^5}C_3 \] \[ {^5}C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \] So, there are 10 arrangements with exactly 3 heads and no two heads are consecutive. 3. Calculate the Probability: \[ P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{128} \] Simplify the fraction: \[ \frac{10}{128} = \frac{5}{64} \] Let me recheck the answer. The calculation seems correct. My result is 5/64, which is option A. The provided answer key says B, which is 5/32 = 10/64. This would imply 20 favorable outcomes. Why would there be 20? The gap method is standard. Let me try to list them. THTHTHT, THTHTTH, THTHTTT, THTHHTT ... this is not practical. The number of ways of choosing $k$ non-consecutive objects from $n$ in a row is $^{n-k+1}C_k$. Here, we are choosing 3 positions for heads out of 7 total positions. $n=7, k=3$. Number of ways = $^{7-3+1}C_3 = ^5C_3 = 10$. This method confirms 10 favorable outcomes. The total outcomes is $2^7=128$. Probability is $10/128 = 5/64$. There is a clear discrepancy between my calculation and the provided answer (5/32). Let's consider if the question could be interpreted differently. "exactly three heads such that no two heads occur consecutively". My interpretation seems to be the only natural one. Maybe the total number of outcomes is not $2^7$? It should be. Maybe the number of favorable outcomes is wrong? The gap method is a standard and reliable technique for this exact type of problem. Let's assume the answer 5/32 is correct. $5/32 = 20/128$. This means there must be 20 favorable outcomes. How could one get 20? Maybe ${^5}C_3 \times 2!$? No, that makes no sense. Is it possible the coin is biased? The problem does not state so. Could it be that the question implies a circular arrangement? "tossed seven times" implies a linear sequence. Let's assume there are 6 tosses instead of 7. Total outcomes = $2^6=64$. 3 heads, 3 tails. Gaps for 3 tails: \_T\_T\_T\_. 4 gaps. Ways = $^4C_3 = 4$. Probability = $4/64 = 1/16$. The calculation $10/128=5/64$ appears to be robust. The answer key is likely incorrect. Step 4: Final Answer
The total number of outcomes is $2^7 = 128$. The number of ways to arrange 3 heads and 4 tails with no two heads consecutive is found by placing the 4 tails first, creating 5 gaps, and choosing 3 of these gaps for the heads, which is ${^5}C_3 = 10$. The probability is $\frac{10}{128} = \frac{5}{64}$. This matches option (A). The provided answer key (B) is likely incorrect.