To find the probability that all 6 flips come up heads, we can use the rule for independent events in probability. The probability of multiple independent events all occurring is the product of their individual probabilities.

Given that the probability of heads on each flip is $P(\text{Heads}) = 0.87$, the probability of getting heads on all 6 flips is:

$P(\text{All 6 heads}) = 0.87 \times 0.87 \times 0.87 \times 0.87 \times 0.87 \times 0.87 = 0.87^6$

Now let's compute this:

$0.87^6 \approx 0.4981$

So, the probability that all 6 flips are heads is approximately 0.4981.

Let me know if you want further details or have any more questions!

Related questions:

1. What is the probability of getting 3 heads in 6 flips of this coin?
2. How would the probability change if the coin had a 0.50 probability of heads?
3. How can the expected number of heads in 6 flips be calculated for this weighted coin?
4. What is the binomial distribution, and how does it apply to this problem?
5. If I flip the coin 10 times, what is the probability that exactly 6 flips come up heads?

Tip:

When dealing with independent events, always remember to multiply their probabilities for the overall outcome.