I’m cramming for a stats test and my brain keeps doing the little spinning wheel whenever “dependent events” pop up. Here’s the specific thing tripping me up:
Bag has 5 red, 3 blue, and 2 green chips (10 total). I draw two chips one after another without replacement.
1) What’s the probability of getting blue then blue? I first wrote (3/10)*(3/10), but then I remembered the no-replacement thing and switched to (3/10)*(2/9). That feels right… but also I keep second-guessing whether I’m mixing up the order or something.
2) What’s the probability of getting at least one blue in two draws? I tried the complement: 1 − P(no blue) = 1 − (7/10)*(6/9). Is this the correct way to handle it when events are dependent, or am I accidentally treating them like independent again?
3) What’s the probability the second draw is blue (regardless of the first)? I tried the law of total probability: P(B2) = P(B2|B1)P(B1) + P(B2|not B1)P(not B1) = (2/9)*(3/10) + (3/9)*(7/10). This seems plausible, but I’m worried I’m double-counting or using the wrong denominators.
Could someone point out where my logic goes wobbly and how to know when to use simple multiplication vs conditional probabilities for dependent events? Any help appreciated!
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
Because draws without replacement are dependent, multiply conditional probabilities: P(B then B) = (3/10)*(2/9) = 1/15; P(at least one blue) = 1 − P(no blue) = 1 − (7/10)*(6/9) = 8/15; P(second is blue) = (2/9)*(3/10) + (3/9)*(7/10) = 27/90 = 3/10 (which also checks out by symmetry, since 3 of 10 chips are blue).
So yes-the complement method is valid here because you’re multiplying the correct conditional probabilities for “no blue.”
That spinning wheel is just your brain’s “conditional probability” hamster sprinting-totally normal! For blue then blue, you nailed it: (3/10)*(2/9) = 6/90 = 1/15, because after a blue leaves the bag, there are only 2 blues among 9 chips left. For “at least one blue,” the complement trick still works perfectly with dependence, as long as the complement is computed conditionally: 1 − (7/10)*(6/9) = 1 − 42/90 = 48/90 = 8/15. For “second draw is blue,” your law-of-total-probability setup is spot on: (2/9)*(3/10) + (3/9)*(7/10) = 6/90 + 21/90 = 27/90 = 3/10; in fact, by symmetry, each position is just as likely to be blue as the overall fraction of blue chips, so 3/10 is a quick mental check. Big picture: the multiplication rule is always P(A and B) = P(A)·P(B|A); independence is the special nap-time where P(B|A) = P(B). The complement rule is always safe too-just compute the complement with the right conditional probabilities. If you like counting, you can also sanity-check by ordered pairs: total 10×9 = 90 outcomes; BB gives 3×2 = 6, no-blue gives 7×6 = 42, and “second is blue” gives 3 choices for second × 9 for first = 27. Hope this helps!
Totally feel you on the spinning wheel – “without replacement” still makes me double-check everything. For blue then blue, you’ve got it: first blue is 3/10, then since one blue is gone and only 9 chips remain, the second-blue chance is 2/9. So P(blue then blue) = (3/10)*(2/9) = 1/15. The (3/10)*(3/10) would be the with-replacement version, so your switch was the right instinct.
For “at least one blue,” the complement route is exactly the right move even when events are dependent. You’re not assuming independence; you’re just computing P(no blue) directly with conditionals: first not-blue is 7/10, then 6/9, so P(no blue) = (7/10)*(6/9) = 7/15, and the complement is 1 − 7/15 = 8/15. Another way (if your brain likes counting) is: number of two-chip draws with no blue is C(7,2), out of all two-chip draws C(10,2), giving 1 − 21/45 = 8/15 as well.
For the second draw being blue, your law-of-total-probability setup is spot on: (2/9)*(3/10) + (3/9)*(7/10) = 3/10. A neat way to see it without any algebra: imagine lining up all 10 chips in a random order before drawing – the chip in position 2 is just as likely to be blue as any random chip, so it’s 3 out of 10. Big picture rule I try to remember (while also second-guessing myself): multiply straight probabilities only if events are independent; when they’re dependent, multiply P(first)*P(second given first). And complements are always fair game – you just compute the complement properly with those same conditionals. It’s like eating two cookies from a jar: the first bite changes what’s left, but the chance the “second bite” is a chocolate chip still matches the overall proportion in the jar to start.