14.3. Exercise 3

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?

Does it mean we take one card out and then take another without putting the first one back?

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?

There would be 51 cards left after drawing the first one.

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

We find the number of hearts in the deck, which is 13, right? The probability for the first heart would be 13/52, and for the second heart, it would be 12/51.

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

We multiply them together since these are dependent events!

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?

We could use the binomial distribution because each toss is an independent trial, right?

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?

n would be 3 since we're tossing three times, p is 0.6, and k would be the number of heads, which can be 0, 1, 2, or 3.

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

We could calculate the probability of getting 0 heads and subtracting that from 1!

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?

We would need to find P(green | not red), which means we can’t just say it's 0.6, right?

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

We need the probabilities of the light being green and not red.

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

If P(red) is 0.4, then the remaining probability of non-red events combined would be 0.6, and we just need to allocate that.

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?