conditional probability (wording, why diff on AND)

Question

A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test? This probability questions is from this website . After reading the explanation, I gained a lot of understanding of conditional probability as a newbie.

but still cannot visualize how P(first and second) is different from P(second |first )? To me, it seems/should be the something(MATH says they are not, but can't full convince myself). While the probability of event first and event second,P(first and second), requires event first happened and event second happened and it is the intersection of venn diagram(is this even correct?). At the same time, P(second |first ) also requires event first happened first and event second happened second.Ok, P(second|first) has order, but still can't see how it different from P(first and second). any tip is appreciated.

On the same website,

Example 1: A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black?"

"The probability of selecting a black marble and then a white marble is 0.34". Would the probability of selecting a white marble and then a black marble be the same as 0.34?

Answer 1

$\Pr(A|B)$ and $\Pr(A\cap B)$ are different, because we're talking about different probability spaces. Suppose, informally, that $A$ means that a student takes calculus, and $B$ means that a student takes physics, and further that, calculus is a prerequisite for physics. Then $\Pr(A|B)=1$: everyone who take physics takes calculus. But surely it needn't be true that $\Pr(A\cap B)=1$. Some students take calculus, but not physics; some take neither.

As I said earlier, we are talking about a different probability space. In this case, the probability space for $\Pr(A|B)$ comprises only the students who takes physics. In the case of $\Pr(A\cap B)$ the probability space comprises all the students. Suppose there are $100$ students, and $60$ take physics. In the first case, the probability measure of a single student is $1/60$. In the second case, the probability of a single student is $1/100$. So, even though we're talking about the same students in both cases, in the first case, we get $\Pr(A|B)=\frac{60}{60}$ and in the second case, we get $\Pr(A\cap B)=\frac{60}{100}.$

Answer 2

Do you agree that P(second |first ) is not the same as P(first |second )? If not, think about raining and cloudy. Given it rains, cloudy is 100%. Give it is cloudy, it might rain.

Now, if you agree that these two probabilities are vastly different, do you see how this leads to a contradiction argument to disprove your intuition?

If you want a more direct argument, try this. The intersection denotes the probability of both first and second happening at the same time, so both events remain to be random. However, by conditioning on first (or second), we ASSUMED that the first (or second) is certain. This is why they are different.