Today we're going to apply what we've learned about probability. Let's start with our first exercise: two cards drawn from a standard deck without replacement. What does that mean, and how do we approach calculating the probability?

Exactly! This affects our total number of possible outcomes. Given a standard deck has 52 cards, how many cards are left after the first draw?

Correct! So if we want to find the probability that both cards are hearts, how would we calculate that?

Great job! Now, how do we combine these probabilities?

Exactly! The probability of both cards being hearts would be (13/52) * (12/51). Let's summarize what we learned: drawing without replacement means changing the sample space, and probability involves multiplication for dependent events.

Now, let’s move on to a slightly different scenario. We have a biased coin with a probability of heads being 0.6. If we toss it 3 times, how can we find the distribution of the number of heads we might get?

Exactly! The binomial probability formula is P(X = k) = (n choose k) * p^k * (1-p)^(n-k). What do you think we would plug in for n, k, and p in this case?

Exactly right! And how would we find the total probability of getting heads at least once?

Perfect! And calculating for k=0, would look like this: P(X = 0) = (3 choose 0) * (0.6^0) * (0.4^3). Now, you all have a good understanding of working with probabilities of biased coins. Let's summarize that for today.

Let's tackle a traffic light scenario. Given that a traffic light is red with a probability of 0.4, what do you think the probability is that the light is green if we know it’s not red?

Exactly! We would use the formula for conditional probability here. What do you think we need to find?

Right! Since P(red) + P(green) + P(yellow) must equal 1, what can we determine here?

Fantastic! Then applying this leads to understanding how to compute P(green | not red) correctly. Always break it into manageable parts! Who can summarize our lesson today?