This is a binomial probability problem, where we are tossing a weighted coin 12 times, and the probability of getting heads is 0.596. We are asked to find the probability of getting heads more than 7 times, i.e., for the number of heads $k > 7$.

The binomial probability formula is:

$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$

Where:

• $n$ is the number of trials (12 tosses),
• $p$ is the probability of success (0.596),
• $X$ is the random variable representing the number of heads,
• $k$ is the number of heads (we want more than 7 heads).

To find the probability of getting more than 7 heads, we need to calculate:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

I'll calculate this for you using binomial probability values for each of these outcomes.The probability of getting more than 7 heads in 12 tosses of a weighted coin with a probability of 0.596 of landing on heads is approximately $0.427$ (rounded to 3 decimal places).

Would you like more details on the steps or any clarifications?

Here are some related questions:

1. How would the probability change if the number of tosses increases?
2. What is the probability of getting exactly 7 heads?
3. How does the probability of tails affect the outcome?
4. How would the distribution look graphically?
5. What happens if the coin is fair, i.e., with $p = 0.5$?

Tip: The binomial distribution becomes more symmetric and approaches a normal distribution as the number of trials increases.