To find the probability of getting at least 1 head when two coins are tossed 1000 times, we can break the problem down step by step.

Given:

• Total tosses: 1000
• Outcomes for each toss (in terms of tails):
  • 0 tails (both heads): 320 times
  • 1 tail (1 head, 1 tail): 352 times
  • 2 tails (both tails): 328 times

Step 1: Find the number of occurrences with at least 1 head

• At least 1 head means we exclude the case where both coins show tails (2 tails).
• The number of times we get at least 1 head is simply the sum of the cases where there are 0 tails (both heads) and 1 tail: $\text{At least 1 head} = \text{0 tails} + \text{1 tail} = 320 + 352 = 672$

Step 2: Calculate the probability

The probability of getting at least 1 head is the ratio of favorable outcomes to total outcomes. So, the probability is: $P(\text{at least 1 head}) = \frac{\text{At least 1 head}}{\text{Total tosses}} = \frac{672}{1000} = 0.672$

Conclusion:

The probability of getting at least 1 head is 0.672 or 67.2%.

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Would you like further details on any part of the calculation, or do you have any other questions related to probability?

Here are some related questions for deeper understanding:

1. What is the probability of getting exactly 1 tail when tossing two coins?
2. How do we compute the probability of getting exactly 2 tails?
3. If the probability of getting heads on each toss was biased, how would that change the probability of getting at least 1 head?
4. How can we generalize this to a situation with more than two coins?
5. What is the expected number of heads in 1000 tosses of two coins?

Tip: When calculating probabilities, always be sure to clearly define the favorable outcomes and total outcomes to avoid confusion.