Bag
Okay, so I need to figure out how to solve this problem about a bag. Let me read it again carefully. The problem says: "Bag contains 5 red marbles, 3 green marbles, and 2 yellow marbles. If you draw 3 marbles without replacement, what is the probability that all three are red?" Alright, so we have a bag with marbles of different colors, and we need to find the probability of drawing three red marbles in a row without putting any back. Let me think step by step.
First, probability problems often involve figuring out the number of favorable outcomes divided by the total number of possible outcomes. Since we're dealing with combinations here (the order in which we draw the marbles might not matter), maybe combinations will be useful. Wait, but actually, since we're drawing without replacement, each draw affects the next. Hmm, so maybe I should approach this using probabilities multiplied together for each draw?
Let me start by recalling the basics. The bag has a total of 5 red marbles, 3 green, and 2 yellow. So that's 5 + 3 + 2 = 10 marbles in total. When we draw the first marble, the probability that it's red is 5 out of 10, which simplifies to 1/2. Then, if we do draw a red marble, there are now 4 red marbles left and the total number of marbles is 9. So the probability of drawing a second red marble is 4/9. Similarly, if we draw another red marble, there will be 3 red left and 8 total marbles. So the third draw would have a probability of 3/8.
So, multiplying these three probabilities together should give the overall probability of drawing three red marbles in a row without replacement. Let me check that. So that would be (5/10) * (4/9) * (3/8). Let me compute that.
https://www.18122.com/url/usdt/usdt2716.jpg
First, 5/10 is 1/2. Then, 1/2 times 4/9 is 4/18, which simplifies to 2/9. Then, 2/9 times 3/8 is 6/72, which simplifies to 1/12. So the probability would be 1/12. Hmm, does that make sense?
Alternatively, maybe I should use combinations. The total number of ways to choose 3 marbles out of 10 is C(10,3), which is 10! / (3! * 7!) = (1098)/(321) = 120. The number of favorable outcomes is the number of ways to choose 3 red marbles out of the 5 available, which is C(5,3). Calculating that: 5! / (3! * 2!) = (54)/(21) = 10. So the probability would be 10/120, which simplifies to 1/12. That's the same result as before. So that's a good check.
Therefore, both methods give the same answer, 1/12. Let me just make sure I didn't make any calculation errors. For the first method: 5/10 is 0.5, 4/9 is approximately 0.444..., multiplying those gives 0.5 * 0.444... 鈮?0.2222, then multiply by 3/8 which is 0.375, so 0.2222 * 0.375 鈮?0.08333..., which is 1/12. Yep, that's correct.
For the combinations method: C(5,3) is 10, C(10,3) is 120, so 10/120 = 1/12. Perfect, same answer. So both approaches confirm that the probability is 1/12.
I think that's solid. Let me just recap. When dealing with probabilities without replacement, you can either multiply the probabilities of each subsequent event (since each draw affects the next) or use combinations to count the number of favorable outcomes over total outcomes. Both methods should give the same result, and in this case, they do. So I feel confident that 1/12 is the correct answer.
The probability of drawing three red marbles without replacement from a bag containing 5 red, 3 green, and 2 yellow marbles is calculated as follows:
Method 1: Sequential Probability
First draw: Probability of red = 5/10 = 1/2.
Second draw: Probability of red = 4/9.
Third draw: Probability of red = 3/8.
Multiplying these probabilities:
[
\frac{5}{10} \times \frac{4}{9} \times \frac{3}{8} = \frac{1}{2} \times \frac{4}{9} \times \frac{3}{8} = \frac{1}{12}
]
Method 2: Combinations
Total ways to choose 3 marbles: (\binom{10}{3} = 120).
Favorable ways (3 red marbles): (\binom{5}{3} = 10).
Probability:
[
\frac{\binom{5}{3}}{\binom{10}{3}} = \frac{10}{120} = \frac{1}{12}
]
Final Answer: (\boxed{\dfrac{1}{12}})
Okay, the user mentioned "BAG". Let me think about what that could refer to.
First, BAG might stand for Business Analysis Group, but I'm not sure. It could also be an acronym in a specific field, like Business Activity Group or something else. Maybe it's related to a company or a project. Alternatively, BAG could be a typo or a shorthand for something else entirely.
I should consider common expansions of BAG. In technology, BAG could be a Backplane Area Group in servers, but that's pretty niche. In medicine, there's the BAG test for blood gas analysis. In logistics, a BAG might be a Baggage Handling System. Or in gaming, there's the game "BAG".
Since the user just wrote "BAG" without context, I need to ask for clarification. Maybe they meant something specific. Let me check if there's a standard meaning for BAG in general. Also, considering the previous interaction was in Chinese, maybe BAG is an abbreviation in Chinese contexts. For example, in Chinese, BAG could be a transliteration or a specific term.
Alternatively, BAG might be part of a larger question that wasn't fully provided. The user might have intended to ask something else but only typed "BAG". Without more context, it's hard to determine the exact meaning. The best approach is to politely ask the user to provide more details or clarify what they need information about regarding BAG.
It seems like you mentioned "BAG," but the context isn't clear. Could you please provide more details or clarify what you're asking about? For example:
BAG might stand for:
Business Analysis Group
Blood Gas Analysis (medical test)
Baggage Handling System (logistics)
Backplane Area Group (technology)
BAG (game) (a mobile game)
Let me know how I can assist! 馃槉
页:
[1]