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急！请教英语数学问题 加声望！Anthony draws s block from a bag with 4 colours of red, green, yellow and blue,he checks its colour, and puts it back. He continues drawing one block at a time and putting it back until he draws the red block.a)What is the probability that Anthony will draw the red block for the first time on his second try?b)Suppose that Anthony does not replace each block after he draws it. Will the probability that he draws a red block on his second try change? Explain. P.S.好像a是3/16。 b是Yes 但是要原因。请用英语，谢谢！
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如果觉得My帖子好请多加点声望！回复: 急！请教英语数学问题你这道题我以前做过，是高中数学B30 中的概率问题。。。只是过了好多年了，早就忘了怎么做这种概率问题。。。。惭愧。。。。不过加拿大数学贼简单，建议你LZ好好看看书，我记得这章里面有几个公式可以代入，就是那3！4！5！那种。。。
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既然来了就留下你的微笑，既然走了就带走你的悲伤回复: 急！请教英语数学问题for question a)probability is independent to the number of trial...therefore, the probability of that his first draw is not red is 3/4, and probability of that his second draw is red is 1/4...you multiply these two probability together, and you will get? (use your brain, or calculator)for question b)probability is independent to the number of trial...therefore, the probability of that his first draw is not red is 3/4 (there are 4 blocks, 1 is red, 3 is not red), and probability of that his second draw is red is 1/3...reason: without replacement, there are 3 blocks, 1 is red, 2 is not red, the probability of 3 choose 1 is 1/3 (the previous question is 4 choose 1, probability of second draw is 1/4)you multiply these two probability together, and you will get? (use your brain, or calculator)
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辛苦十几年，你也不过是红警里500块一个的工程师，一条狗咬死一片的那种。。。CBC Radio 2 ClassicalMy RSS Feeds List你这道题我以前做过，是高中数学B30 中的概率问题。。。只是过了好多年了，早就忘了怎么做这种概率问题。。。。惭愧。。。。不过加拿大数学贼简单，建议你LZ好好看看书，我记得这章里面有几个公式可以代入，就是那3！4！5！那种。。。点击展开...这道题还没有难到要用factorial。。。以后学到binomial和hypergeomatric才会多用到factorial...而且以后如果binomial太多terms的话，根据条件可以用exponential或者normal destribution去大约估算binomial的概率（用continuous估算discrete），方便很多。。。评论
辛苦十几年，你也不过是红警里500块一个的工程师，一条狗咬死一片的那种。。。CBC Radio 2 ClassicalMy RSS Feeds List这道题还没有难到要用factorial。。。以后学到binomial和hypergeomatric才会多用到factorial...而且以后如果binomial太多terms的话，根据条件可以用exponential或者normal destribution去大约估算binomial的概率（用continuous估算discrete），方便很多。。。点击展开...你是数学狂人吧吧。。。。太牛B了评论
既然来了就留下你的微笑，既然走了就带走你的悲伤你是数学狂人吧吧。。。。太牛B了点击展开...。。。。。。LZ问的问题，在中国这些是高中数学的吧。。。 后面那些数理统计的东西我在Kwantlen学的，大二数学课，老师教得不错。。。
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辛苦十几年，你也不过是红警里500块一个的工程师，一条狗咬死一片的那种。。。CBC Radio 2 ClassicalMy RSS Feeds Listfor question a)probability is independent to the number of trial...therefore, the probability of that his first draw is not red is 3/4, and probability of that his second draw is red is 1/4...you multiply these two probability together, and you will get? (use your brain, or calculator)for question b)probability is independent to the number of trial...therefore, the probability of that his first draw is not red is 3/4 (there are 4 blocks, 1 is red, 3 is not red), and probability of that his second draw is red is 1/3...reason: without replacement, there are 3 blocks, 1 is red, 2 is not red, the probability of 3 choose 1 is 1/3 (the previous question is 4 choose 1, probability of second draw is 1/4)you multiply these two probability together, and you will get? (use your brain, or calculator)点击展开...b比较好理解，a没明白为什么要用两个概率相乘？？？评论
既然来了就留下你的微笑，既然走了就带走你的悲伤。。。。。。LZ问的问题，在中国这些是高中数学的吧。。。 后面那些数理统计的东西我在Kwantlen学的，大二数学课，老师教得不错。。。点击展开...我在国内只上了高1，而且学的很烂，基本上是什么都不知道啊。。惭愧
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既然来了就留下你的微笑，既然走了就带走你的悲伤b比较好理解，a没明白为什么要用两个概率相乘？？？点击展开...在第一种情况（第一次抽不到红色）之后，才会有第二种情况（第二次抽到红色）出现。。。 所以：第一次就抽到红色的几率是；1/4第二次才抽到红色的几率是；3/4 * 1/4第三次才抽到红色的几率是；3/4 * 3/4 * 1/4第四次才抽到红色的几率是；3/4 * 3/4 * 3/4 * 1/4etc.
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辛苦十几年，你也不过是红警里500块一个的工程师，一条狗咬死一片的那种。。。CBC Radio 2 ClassicalMy RSS Feeds List我在国内只上了高1，而且学的很烂，基本上是什么都不知道啊。。惭愧点击展开...呵呵，没关系。。。如果以后不需要数学的话，学了都没什么用。。。省点脑子，想想怎么泡妞~评论
辛苦十几年，你也不过是红警里500块一个的工程师，一条狗咬死一片的那种。。。CBC Radio 2 ClassicalMy RSS Feeds List回复: 急！请教英语数学问题。评论
有一种情怀,淡如轻烟,却能长久而执著地散发着暖人的温馨;有一种感动,默无声息,却能让我们为之心酸并为之泪流满面.回复: 急！请教英语数学问题多谢你们！
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