In this problem, Greg states that pulling a red marble and keeping it, then pulling a blue marble is considered a case of independent events. Can somebody elaborate further on why? How can they be independent if the probability for pull 2 is inherently changed? Greg says to treat it like a new bag of marbles but I don’t understand how they’d still be considered independent on a GRE question?
Hi, thanks for your reply. I understand the equal probabilities between pulling a certain ball 1st vs. 5th, etc. From reviewing the thread, I am not quite sure that answers my question as to how the events are considered independent. Is that fact that the probability doesn’t change proof of the independence of events, or am I missing some sort of connection? Thanks!
Yes, events A and B are independent if P(B given A occurred) = P(B) (A happening does not change B’s likelihood → B is exactly as likely to happen whether A happens or not). The same goes if you swap the letters A and B above in the definition (i.e. B happening does not change A’s likelihood of happening).
If the probability of one event changes when the other occurs, then those events are not independent (aka dependent).
Thanks for your response. The technicality I am hung up on is this - in this case, isn’t the probability of one event changing when the other occurs, given that the total available number of outcomes is changing? If A occurs, probability of B changes from 5/10 to 5/9, correct?
Those should not be independent events because the probability changes. Selecting a red increases the likelihood of a blue on 2nd try. Where is that slide? We should correct it
If you did replace the marbles after drawing it, then those events would be independent.
This is under the independent events video on PrepSwift, towards the end of the video.
Hi it seems this is still not corrected. The Prepswift video is “Independent Events”. Around the 3m minute mark is when the problematic example begins. In it, Greg says pulling with replacement is independent (correct), and he further argues that pulling without replacement is also independent (incorrect – this changes the probability of the second draw). This is incorrect, no?
The two events are independent. To show this, let
- event A be the probability of getting the red on the first pull
- event B be the probability of getting the blue on the second pull
The probability of both events happening, P(A \cap B) is
P(A \cap B) = \frac{5}{10} \times \frac{5}{9} = P(A) \times P(B)
Note in particular that P(A \cap B) = P(A) \times P(B), which suffices as a condition for independence. It can be similarly shown that P(A | B) = P(A).
But how do we relate this? One way to think of it is for event B, the first pull must have happened. Once the first pull has happened, we’re in a new state. And this new state is detached from whatever happened in event A (which is what Greg implies with his example of a new bag).
An example that would not be independent is the event of picking two coloured balls. Can you show why? Hint: the reason the first example is independent is that it’s essentially a specific case (red then blue immediately).