两道英文数学题第一题：involving the rolling of a fair die. Assume that

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两道英文数学题
第一题：
involving
the rolling of a fair die. Assume that the die is rolled 12 times,
and the roll is called a success if the result is in

{1,2}
What is the probability that there are exactly 4 successes or
exactly 4 failures in the 12 rolls?
第二题：involving
Tom's class attendance. Assume that he attends randomly with probability 0.55
and that each decision is independent of previous attendance, so that the
process can be viewed as a Bernoulli process.
What is the probability that he attends at least 7 of 10 classes
given that he attends at least 2 but not all 10 classes?
求呐.希望上次那个人再次出现拯救我!

先是第一题~【我不是“上次拯救你的那个人”.】
1、大概翻译一下,就是一个均匀的骰子投掷12次,当结果是1或2的时候是成功,反之失败.求确切投到4次成功或确切的4次失败的概率.
那么,对单次投掷来讲,成功的概率p=2/6=1/3,失败q=2/3；
而4次成功和4次失败没有交集,所以分别求出来概率然后相加就好了
4次成功概率=C(4,12)*(1/3)^4*(2/3)^8,C(4,12)是12个里选4个,C(4,12)=(12*11*10*9)/(4*3*2)
4此失败概率=C(8,12)*(1/3)^8*(2/3)^4.
然后两个加一起~
大概是这样的,就是那个n次独立重复试验好像是……不过好久没做过这样的题了~有不对的地方大家尽快帮忙纠正啦……
那我先去看看第二题
恩,第二题比刚才那个稍微复杂一点儿,不过方法是一样的.大概是酱紫：
2、翻译：Tom小朋友去选课- -#他是随机选择的,选的概率是p=0.55,他做的每个决定都是独立的,和前面没有关系,也就是说是一个伯努利过程.（也就是n次独立重复事件）问：在已知他至少选了10门里的2门但是不选全的情况下,选了至少7门的概率.
这是一条件概率的问题.所以应该是：
Pr(所求)=Pr(至少选7门|至少2门但是不选全)=Pr(选了7or8or9)/Pr(选了2or3or4or...or9门)
这个说实话有点儿麻烦,选几门之间的事件都是独立且不重叠的,所以要加起来.
给个公式吧,比如是选7门的概率,就是在10次事件中发生7次
Pr(7)=C(7,10)*p^7*(1-p)^3,p=0.55
然后加加除除就好了~

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