Calculating the Probability of Getting Exactly 2 Heads in 8 Coin Tosses with at Least One Head
When tossing a fair coin 8 times, the question arises: what is the probability of getting exactly 2 heads, given that at least one head has appeared? To answer this, we need to explore the concepts of binomial distribution, conditional probability, and combinatorial mathematics.
Understanding the Combinations and Probabilities
The total number of possible outcomes when a fair coin is tossed 8 times is (2^8 256). This is because each toss has 2 possible outcomes (heads or tails), and multiplying these outcomes for 8 tosses gives us 256 combinations.
The number of combinations that result in exactly 2 heads can be determined using the binomial coefficient, which is calculated as follows:
(C(8, 2) frac{8!}{2!(8-2)!} 28)
Now, let's consider the combinations with at least one head. There is one combination with no heads (all tails), and there are 8 combinations with exactly one head (one head and seven tails). Therefore, the number of combinations with at least two heads is (256 - 1 - 8 247).
Calculating the Probability
The probability of getting at least two heads is given by:
[ P(X geq 2) frac{247}{256} approx 0.965... ]The probability of getting exactly 2 heads, given that at least one head has appeared, can be calculated using the conditional probability formula. The formula we use is:
[ P(X 2 mid X geq 1) frac{P(X 2 text{ and } X geq 1)}{P(X geq 1)} ]Since (P(X 2 text{ and } X geq 1) P(X 2)) and (P(X geq 1) 1 - P(X 0)), we have:
[ P(X 2) frac{28}{256} frac{7}{64} ] [ P(X geq 1) 1 - frac{1}{256} frac{255}{256} ]Substituting these into the conditional probability formula, we get:
[ P(X 2 mid X geq 1) frac{frac{7}{64}}{frac{255}{256}} frac{28}{255} approx 0.1098 ]Checking with a Brute Force Approach Using J Programming Language
A brute force approach using the J programming language can also be used to verify this result. In this method, we generate all possible outcomes of 8 coin tosses (256 outcomes) and count how many have exactly 2 heads. Then, we divide this count by the total number of possible outcomes:
[ text{Probability} frac{text{Number of outcomes with exactly 2 heads}}{text{Total number of outcomes}} frac{56}{256} frac{7}{32} 0.21875 ]Conclusion
In conclusion, the probability of getting exactly 2 heads in 8 coin tosses, given that at least one head has appeared, is approximately 0.1098 or 10.98%. This can be calculated using the principles of binomial distribution and conditional probability, and verified through a brute force approach using programming languages like J.
Key points to remember:
- The total number of possible outcomes is (2^8 256). - The number of combinations with 2 heads is (binom{8}{2} 28). - The probability of exactly 2 heads given at least one head is approximately 0.1098.This calculation showcases the power of combinatorial mathematics and probability theory in understanding and solving real-world problems.