I’m prepping for a test and my brain keeps trying to sneak marbles back into the bag when they’re not invited. Suppose I’m drawing two marbles from a bag without replacement. I know this is a “dependent events” situation, but I get tangled about how to think about it. Why exactly does the chance of the second color change just because of what happened first? Like, if the first marble is red, I get that the bag shrinks and the vibe changes… but is there a simple way to keep that straight in my head without accidentally using the independent-events mindset?
Also, here’s the part that really scrambles my cereal: if I draw the first marble and don’t look at it (I just set it aside like a mysterious marble of destiny), is the second draw still dependent? And follow-up: if the problem says I drew a red first but I put it back before drawing again, does that make the events independent again even though I “know” something about the first draw?
Could someone explain how to recognize these dependent situations at a glance and why the second probability actually shifts? Bonus points if you have a little mental trick for deciding when I should treat draws as dependent vs independent.
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3 Responses
I have the same “oops, I mentally put the marble back” habit! The quick rule is: ask whether the pool you’re drawing from changes for the second draw, or whether you’re conditioning on the first result. If the composition changes (no replacement), the second probability shifts. If the composition resets (with replacement), it stays the same. Example: suppose the bag has 3 red and 2 blue (5 total). Without replacement, if the first is red, then the chance the second is red becomes (3−1)/(5−1) = 2/4 = 1/2. If the first is blue, it becomes 3/4. Different first outcomes lead to different second-draw probabilities, so the events are dependent.
Now for the “mysterious marble of destiny” that you don’t look at: the second draw’s standalone probability of red is still 3/5. You can see this by splitting on what the first was: P(second red) = P(second red | first red)·P(first red) + P(second red | first blue)·P(first blue) = (2/4)(3/5) + (3/4)(2/5) = 3/5. So if you don’t peek, you should still use 3/5 for the second draw. But the events are still dependent, because if you did peek, your probability for the second would change to 1/2 or 3/4. Your ignorance doesn’t put the marble back; it just means you aren’t conditioning on the first result.
If you draw a red and then put it back before drawing again (with replacement), the bag goes back to 3 red out of 5, so P(second red | first red) = 3/5 = P(second red). That equality is the hallmark of independence. Mental trick: dependent if either (a) the first outcome changes the pool/mechanism for the second, or (b) knowing the first would change your probability for the second. Independent if neither happens.
I think the second draw only really changes if you actually peek at the first-like grabbing a cookie without looking doesn’t change the snack bowl’s vibe-so from 3 red and 1 blue, not looking keeps the next-red chance at 3/4 (but if you saw a red first it drops to 2/3), and if you pop it back in, I’m pretty sure they’re independent again.
Think of the bag like a cookie jar that actually loses a cookie when you take one. If there are R red and B blue marbles, then before any draw the chance of red is R/(R+B). After you actually see the first marble, the jar’s contents have changed, so the conditional chance for the second draw shifts: P(second red | first red) = (R−1)/(R+B−1) and P(second red | first blue) = R/(R+B−1). That’s the whole “dependence” in action: the second probability is conditioned on what happened first, and the pool you’re drawing from is different. A handy mental cue: if the pool changes and stays changed, the next probability should change too. Khan Academy has a nice quick refresher on this idea here: https://www.khanacademy.org/math/statistics-probability/probability-library/basic-theoretical-probability/v/probability-with-and-without-replacement
Now, the sneaky part: if you draw the first marble and don’t look, your probability for the second draw being red is still R/(R+B). You’ve physically changed the bag, but because you didn’t condition on what came out, the ups and downs balance out: P(R2) = P(R2|R1)P(R1) + P(R2|B1)P(B1) = (R−1)/(R+B−1)·R/(R+B) + R/(R+B−1)·B/(R+B) = R/(R+B). The events are still dependent overall (if you were to peek, you’d update to one of those shifted fractions), but unconditionally the second-draw probability matches the original proportion. And if you do replace the first marble before drawing again, the pool resets, so P(second red | first red) = R/(R+B) again-now the events are genuinely independent, regardless of what you know about the first draw.
Quick rule of thumb: ask “Does what happened earlier change the pool I’m drawing from for the next step?” If yes and it stays changed, treat them as dependent. If no (you replace or reset), treat them as independent. Want to test it with numbers? Say the bag has 5 red and 7 blue: what’s P(second red) if you don’t look at the first, and what are P(second red | first red) and P(second red | first blue)?