Probability of Multiple Weighted Coin Flips

Question

I would like to calculate the probabilities of the outcomes of three weighted coins being flipped. I believe what I am looking for is a Poisson Binomial Distribution.

I am having trouble verifying/interpreting the results that I am finding on an online calculator.

Edit: Order does not matter in this question - the below is a table of the sums of outcomes.

+---------+---------+-------------+
| Outcome | Heads P | Probability |
+---------+---------+-------------+
| 3 heads |     .75 | .421875??   |
| 2 heads |     .75 | ??          |
| 1 heads |     .75 | ??          |
| 0 heads |     .75 | ??          |
|         |         |  (Sum = 1)  |
+---------+---------+-------------+

The .42 is calcualted for X>=3, but since there are only 3 flips it cannot be any greater. An alternate calculator provides a much lower answer, .03, which seems too low.

1. Is a Poisson binomial distribution the correct calculation for this answer? (X=3 trials, .75 success)
2. How would I find the probability of 2 out of 3 heads, 1 out of 3 heads, and no heads?

Thank you for taking the time to explain what I might be missing here.

Answer 1

In a problem like this, where there are only 8 possible outcomes, sometimes it is simpler just to list them all and calculate the probability of each.

The probability of HHH is 0.75 ^ 3 which is about 0.4219.
The probability of TTT is 0.25 ^ 3 which is about 0.0156.

The probability of HHT, HTH, and THH is 0.1406 each (.75^2 * .25) so 0.4218 total.

The probability of HTT, THT, and TTH is 0.0469 each (.25^2 * .75) so 0.1407 total.

All of these 8 possible outcomes sum up to probability 1 (discarding roundoff error).

Just as a "side note", mathematics has a powerful set of tools, but for simple problems, all you need are simple tools. Match the tool the the problem and you will get results.

Answer 2

If the coins all have the same bias (0.75 for showing heads), then the distribution for the count of heads among three flips is Binomial.

$$X\sim\mathcal{Binomial}(3,0.75)\quad\iff\quad\mathsf P(X{=}x) ~=~ \binom{3}{x}\,0.75^x\,0.25^{3-x}\,\mathbf 1_{x\in\{0,1,2,3\}}$$

(Poisson Binomial is when the coins each have a distinct bias).