help!

cliffca

5．    生4个小孩，2男2女的概率

象这种题,还有投硬币的题,应该怎么算?

sunnY_yueliu

I was confused before when I saw the answer such as: C(4,2)/2^4 = 6/16.

Now I finaly figure out the above approach and would like to share with you:

There are two analysis approaches:

Choice approach:consider the choice of each item(each baby or each dice). each baby have two choices (boy or girl)
So the total choices = 2^4 = 16
So the 2B and 2G choices = First Baby's choice * (Second baby chose the same as first baby + second baby chose the different from first baby * the third baby's choice) = C(2,1) * (C(1,1) + C(1,1) * C (2,1)) = 2 * (1+2)=6

Result approch: consider the final result as whole. (how many are boyes and how many are girles)
So the total results = zero girl + 1 girl + 2 girls + 3 girls + 4 girls = C(4,0) + C(4,1) + C(4,2) + C(4,3) + C(4,4) = 1 + 4 + 6 +4 +1 =16
So the 2B and 2G results= C(4,2)= 6

It is easy to get the total number from the choice approach and is easy to get the spec number from result approach. So once you fully understand the issue, you can use the mixed approach like those "Da Niu" C(4,2)/2^4

Hope I am not making you more confused.

may

掷一个均匀硬币2N次，求出现正面K次的概率。
C2n，k(1/2) ^2n 独立重复试验。如果在一次试验中某事件发生的概率是P，那么在n次独立重复试验中这个事件恰好发生K次的概率为Pn（K）＝Cn,kP^k (1－P)^( n-k)
（一夫妇生四孩子，问生2男2女的情况之几率；每次生男女概率相同，1/2,如抛硬币问题（抛四次，2次朝上）,即C42(1/2) ^4＝3/8
一个COIN ，TOSS 三次，三次都有同样面(正反都有可能)的概率：(1/4) 两次：1/2
我想在考试中再遇到同类题，你一定错不了了。错了打屁股。

cliffca

thank u so much!