几道JJ，大家快来棒棒我！多谢！

快要考了还不会呢！惨！

1，关于两个事件发生的概率问题，e.g. 两个事件A，B概率分别是0.5和0.4，问A不发生，B也不发生的概率范围？

什么时候是0.1~0.5?什么时候是0.02啊？

2，2个盒子，一个4个apple，2个orrange,另一个盒子3个apple，5个orrange，问随即从盒子里取， 一个是apple, 另一个是orrange的概率？

请问：如果不说什么先后顺序，是不是就是C(7，1)*C(7，1)/(14*13)； 要是说先从第一个盒子里去一个apple,然后从第二个盒子里去一个orrange的概率，那就是：C(4，1)*C(5，1)/[C(6，1)*C(8，1)]？

3， 2个sit，S和T，是不是T的标准方差比S的大？(1)S都是25的倍数；(2)T都是5的倍数

答案是e，请问根据什么判断标准方差的大小呢？

多谢

let's say we have 10 numbers: 2, 4, 8, 12, 16, 1, 3, 5, 7, 9, if we randomly pick 1 number, let a be the probability that the number we picked is even, let b be the probability that the number we picked is divisible by 4, then a = .5(2, 4, 8, 12, 16), b = .4(4, 8, 12, 16); so what's the probability a and b won't happen? since a contains b then this is simply 1 - a = .5, this is the upper bound of the range;

let's say another 10 numbers: 2, 4, 6, 8, 10, 3, 6, 9, 12, 7, if we randomly pick 1 number, let a denote the probability that the number we picked is even (this time a = .5), let be denote the probability that the number is disable by 3(b = .4); this time around, the chance of a and b not happen at the same time is: 1 - .5 - .4 = .1 since they are independent of each other, and this is the lower bound of the range;

thus the range for this problem is .1 ~.5

as for the .2 case you asked for:

another 10 number set: 2, 4, 6, 8, 10, 9, 15, 21, 5, 7, we are going to pick another random number, let a be the probability of us picking an even number (.5), and b be the probability of using picking a number that's divisible by 3(.4), so what's the probability that the number we picked is neither even or disable by 3? the answer is: 1 - a - b + a u b(means a and b, in this case, it's .1, since a and b both share the number 6, and 1 number out of 10 is .1) = 1 - .5 - .4 + .1 = .2(leaves 5 and 7 to be the only choices)

I hope this helps...

S = standard deviation Σ = sum of X = individual score M = mean of all scores n = sample size (number of scores)

i think this is what you are talking about; i didn't take math beyond highschool level in China, so some of the terms I'm just not sure how to translate, if this is what you are talking about, the key is the mean of the set, u don't have to know each number sometimes, but the mean is very important;

for example if we use ur qustion:

first thing first, u need to know that if you can solve this problem, you will need BOTH 1) and 2), since you will need to know about both sets in order to draw the conclusion, so we know from 1) if s is in S, then s = 25 * n, and 2) if t is in T, then t = 5 * m, where n and m are positive integers...

now, we have no idea what the mean is for each set, it is enough to draw the conclusion that we have no idea what the sds are for each set (sometimes, though, if one of the premises tells you something about the means of each sets, maybe it is enough to know the sds)...

I don't kown that well about sds since it's been a while since i've done them, so please, if i made any error on anything, call me out...

I'm not very good with combos, and usually i do the stupid and safe way which is:

COUNT THEM OUT BY HAND, 1, 2, 3, 4, 5, 6...

.........

JUST KIDDING

.........

.........

...as ya'll right now probably shaking your heads and laughing at my stupid sense of humor, here's how i would try to solve this question:

1) if order doesn't matter:

how many ways can we actually get 1 apple and 1 orange from two boxes?

well, in general there are two ways: (1) get 1 apple from the 1st box and 1 orange from the 2nd box, (2) get 1 orange from the first box and 1 apple from the 2nd box

and let's do that: (1) c(4, 1)/c(6, 1) * c(5, 1)/c(8, 1), (2) c(2, 1)/c(6, 1) * c(3, 1)/c(8, 1)

and we add them together: we have (4 * 5 + 2 * 3) / (6 * 8) = 26 / 48, which is about 54%, sounds reasonable...

in order to check this, we can a similiar problem with a smaller sample size, for example: if we have the 1st box has 1 apple and 1 orange, and 2nd box has 1 apple and 1 orange, then we want to know the same thing we would have (1 * 1 + 1 * 1)/(2 * 2) = 1/2 = 50%, since the only outcomes are: (1) 1st box apple, 2nd box orange, (2) 1st box apple, 2nd box apple (3) 1st box orange, 2nd box apple, (4) 1st box orange, 2nd box orange...

2) if order does matter:

you actually got it right...

谢谢tiger GG，现在明白独立和包含了。但是，比如下边这题，说A发生的概率是0.5，B发生的概率是0.4，A和B不发生时C发生，问C的概率范围。(考答案是0.01;0.8;0.9)

还有一个：F发生的概率是50%，T是40%，问两个都不发生的概率，参考答案是：0.3

还有：两个事件相互独立，概率各是P，问A发生B不发生的概率， 参考答案是P-P^2

前面两个是不是就按照你讲的独立和包含的方法做？就是如果先考虑A，B独立，两个都不发生的概率是：1-0.5-0.4=0.1;

然后在考虑两个包含的概率：1-0.5=0.5

但是我对你说的第三种情况还是不太明白，因为你举了具体数字，所以我知道A并B是多少，但是象上面这几题，怎么知道AUB是0.1还是0.2呢？

以下是引用snowflower在2004-8-16 6:57:00的发言：

谢谢tiger GG，现在明白独立和包含了。但是，比如下边这题，说A发生的概率是0.5，B发生的概率是0.4，A和B不发生时C发生，问C的概率范围。(考答案是0.01;0.8;0.9)

还有一个：F发生的概率是50%，T是40%，问两个都不发生的概率，参考答案是：0.3

还有：两个事件相互独立，概率各是P，问A发生B不发生的概率， 参考答案是P-P^2

前面两个是不是就按照你讲的独立和包含的方法做？就是如果先考虑A，B独立，两个都不发生的概率是：1-0.5-0.4=0.1;

然后在考虑两个包含的概率：1-0.5=0.5

但是我对你说的第三种情况还是不太明白，因为你举了具体数字，所以我知道A并B是多少，但是象上面这几题，怎么知道AUB是0.1还是0.2呢？

还有：两个事件相互独立，概率各是P，问A发生B不发生的概率， 参考答案是P-P^2

let's take a look at this first;

so we have a = p, b = p, and they are independent, so what's the probability a happens and b doesn't happen?

if a and b are independent, then a and ~b are also independent.(take my word for it, if u don't, ask me to prove it and i will)

and Pr(ab) = Pr(a) * Pr(b) so if we want to know Pr(a~b) and a and ~b are independent, we have Pr(a) * Pr(~b), since Pr(a) = Pr(b) = p, it follows that Pr(b) = 1 - p, thus Pr(a~b) = p * (1 - p) = p - p^2...

as for 说A发生的概率是0.5，B发生的概率是0.4，A和B不发生时C发生，问C的概率范围。(考答案是0.01;0.8;0.9), i don't get that answer, does that mean .01/.8/.9 or something? sorry...

F发生的概率是50%，T是40%，问两个都不发生的概率，参考答案是：0.3

i would know how to solve this, im pretty sure more information must be given in order to solve it...

i hope im helping...

因为是JJ，所以可能有不全的地方，再加上我基础太差，所以放到一起看就晕了！:p

但是我对你说的第三种情况还是不太明白，因为你举了具体数字，所以我知道A并B是多少，但是象上面这几题，怎么知道AUB是0.1还是0.2呢？

噢，知道了，多谢tigerGG。

这会儿终於明白了！