Final answer:
The probability of getting more than 5 heads in 10 coin tosses can be calculated using the binomial distribution. Each of the possibilities (6, 7, 8, 9, and 10 heads) needs to be individually calculated and then summed up to find the total probability. This process illustrates the concept of the law of large numbers.
Step-by-step explanation:
To answer the question about the probability of getting more than 5 heads in 10 coin tosses, we can utilize the binomial distribution model. Since each coin toss is an independent event and the coin is fair, we have a probability (p) of getting heads on a single toss equal to 0.5, and consequently, the probability of tails (q) is also 0.5. The number of trials (n) is equal to 10 since the coin is tossed 10 times.
The probability of getting exactly k heads in n tosses of a fair coin, where the probability of heads is p (0.5 in this case), is given by the formula for the binomial probability:
P(X = k) = C(n, k) * p^k * q^(n-k)
Where C(n, k) is the combination of n things taken k at a time.
Since we are looking for the probability of getting more than 5 heads, we need to sum the probabilities for 6, 7, 8, 9, and 10 heads:
- P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
We can calculate each of these probabilities using the binomial formula and then add them together to find the total probability of getting more than 5 heads in 10 flips of a fair coin.
This kind of exercise helps to reinforce the concepts of theoretical and empirical probabilities, and it illustrates the law of large numbers, which says that as the number of trials increases, the experimental probability tends to approach the theoretical probability.