So I initially made the mistake of thinking this question was asking for the probability of pulling out a blue marble on the third trial given one is doing three trials, which made me think this involved combinatorics.
Now that this confusion has been cleared up I am still confused. It seems like there are 4 ways to pull a blue marble on the third try:
RRB = (3/10)(2/9)(4/8)
GGB = (3/10)(2/9)(4/8)
RGB = (3/10)(3/9)(4/8)
GRB = (3/10)(3/9)(4/8)
Since RRB=GGB and RGB=GRB, we have 2((3/10)(2/9)(4/8)) + 2((3/10)(3/9)(4/8))=120/720.
Why is this wrong?
Hmmm I suppose I misread “pulling on third try” as meaning “pulling only on third try.” i will try include this in my calculation. Thanks.
OK so BBB would have proability (4/10)(3/9)(2/8)=24/720. Adding this to the previous probabilities gives us 144/720=0.2.
What about BGB and GBB?
Hint: look at “Introduction to Probability” on PrepSwift.
I was about to post about this problem as well since I was struggling with it too. To clarify, we should find all combinations in which blue is chosen third, calculate the probability of each, and add those probabilities to confirm that the probability of choosing blue third will always be 2/5 regardless of replacement? I did this and got 288/720 (2/5) as my answer, but now I am wondering why the combinations after the first three picks don’t affect this probability too. For example BBBRRRGGGB and BBBRGRGRGB, among the many additional endings to this. Is this because the problem implies picking only three times, not ten?
I’d recommend seeing the video again, and
Yes. The fourth item onwards is out-of-scope to the question.
