And then times 0.3, 30% chance of missing the third, times a 70% chance of making the fourth, times a 30% chance for each of the next two misses. If you wanted this exact circumstance, this exact circumstance, this is once again going to be 0.7, if you just rearrange the order in which you're multiplying, this is going to be 0.7 squared times 0.3 to the fourth power. So for any one of these particular ways to get exactly two scores in six attempts, the probability is going to be this. So the probability of getting exactly two scores in six attempts, well it's going to be any one of these probabilities times the number of ways you can get six scores, times the number of ways you can get two scores in six attempts. Well, if you have, out of six attempts, you're choosing two of them to have scores, how many ways are there? Well, as you can imagine, this is a combinatorics problem. So you could write this as, you could write this, and let me see how I could, you're gonna take six attempts, so you could write this as six choose, or we're trying to, if you're picking from six things, your six attempts, and you're picking two of them, or two of them are going to be, need to be made if you want to meet these circumstances.