Question

The probability that exactly 3 heads appear in six tosses of an unbiased coin, given that the first three tosses resulted in 2 or more heads, is:

• A. \( \frac{3}{16} \)
• B. \( \frac{5}{16} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{9}{16} \)

Hint:

When calculating conditional probability, first determine the probability of the given condition occurring, then use Bayes' theorem or probability ratios to find the required result.

Answer 1

The Correct Option is B

We are given that a fair coin is tossed 6 times, and we need to find the conditional probability that exactly 3 heads appear given that the first three tosses resulted in at least 2 heads. 

Step 1: Define total probability space 
Each toss of the coin is independent with two outcomes: head (H) or tail (T). The probability of any specific sequence of 6 tosses occurring is: \[ \left(\frac{1}{2}\right)^6 = \frac{1}{64} \] 

Step 2: Compute probability of the given condition 
The probability that the first three tosses result in at least 2 heads can be computed by considering the cases: 1. \( (H, H, H) \) 2. \( (H, H, T) \) 3. \( (H, T, H) \) 4. \( (T, H, H) \) Each of these cases follows a binomial probability distribution: \[ P(\text{At least 2 heads in first 3 tosses}) = P(2H) + P(3H) \] Using the binomial formula: \[ P(2H) = \binom{3}{2} \left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8} \] \[ P(3H) = \binom{3}{3} \left(\frac{1}{2}\right)^3 = 1 \times \frac{1}{8} = \frac{1}{8} \] \[ P(\text{At least 2 heads in first 3 tosses}) = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \] 

Step 3: Compute probability of exactly 3 heads given the condition 
We need to find \( P(X = 3 | \text{At least 2 heads in first 3 tosses}) \), which is given by: \[ P(X = 3 \cap A) / P(A) \] where \( A \) is the event that at least 2 heads occur in the first 3 tosses. For exactly 3 heads in 6 tosses, given that at least 2 heads occurred in the first 3 tosses, the remaining 3 tosses must contribute either 0 or 1 additional head. This follows the binomial probability: \[ P(3H \text{ in 6 tosses} | A) = \frac{5}{16} \] 

Step 4: Conclusion 
Thus, the correct answer is: \[ \mathbf{\frac{5}{16}} \]