实题操作
1. 三个人独立地去破译一份密码，每人能独立译出这份密码的概率分别为1/5， 1/3， 1/4。则这份密码被译出的概率为（3/5）。
def success(): p = 1/5,1/3,1/4 t = 1 for i in p: t *= 1-i return 1-t print(f'成功概率：{success():.3f}')
2. 甲、乙、丙三人向同一飞机射击，假设他们的命中率都是0.4；又若只有一人命中时，飞机坠毁的概率为 0.2；恰有两人命中时，飞机坠毁的概率为 0.6；若三人同时命中，飞机必坠毁。求飞机坠毁的概率为（202/625）。
p1 = [1-0.4,0.4] # 不中和击种的概率 p2 = [0,0.2,0.6,1.0] # 不同人数击中的坠机概率 p0 = 0 # 初始总概率 for i in range(2): for j in range(2): for k in range(2): t = p1[i] * p1[j] * p1[k] * p2[i+j+k] p0 += t print(f'{i} {j} {k} : {p1[i]} * {p1[j]} * {p1[k]} * {p2[i+j+k]:.1f} = {t:.4f}') import fractions print(f'\nTotal : {fractions.Fraction(str(0.3232))}') ''' 0 0 0 : 0.6 * 0.6 * 0.6 * 0.0 = 0.0000 0 0 1 : 0.6 * 0.6 * 0.4 * 0.2 = 0.0288 0 1 0 : 0.6 * 0.4 * 0.6 * 0.2 = 0.0288 0 1 1 : 0.6 * 0.4 * 0.4 * 0.6 = 0.0576 1 0 0 : 0.4 * 0.6 * 0.6 * 0.2 = 0.0288 1 0 1 : 0.4 * 0.6 * 0.4 * 0.6 = 0.0576 1 1 0 : 0.4 * 0.4 * 0.6 * 0.6 = 0.0576 1 1 1 : 0.4 * 0.4 * 0.4 * 1.0 = 0.0640 Total : 202/625 '''
3. 甲、乙、丙三人向同一飞机射击，假设他们的命中率分别是：0.4, 0.5, 0.7；又若只有一人命中时，飞机坠毁的概率为 0.2；恰有两人命中时，飞机坠毁的概率为 0.6；若三人同时命中，飞机必坠毁。求飞机坠毁的概率为（229/500）。
继上题，只是3人的命中率不同；代码稍作修改即可：
p1 = [[1-0.4,0.4],[1-0.5,0.5],[1-0.7,0.7]] p2 = [0,0.2,0.6,1.0] # 不同人数击中的坠机概率 p0 = 0 # 初始总概率 for i in range(2): for j in range(2): for k in range(2): t = p1[0][i] * p1[1][j] * p1[2][k] * p2[i+j+k] p0 += t print(f'{i} {j} {k} : {p1[0][i]:.1f} * {p1[1][j]:.1f} * {p1[2][k]:.1f} * {p2[i+j+k]:.1f} = {t:.4f}') import fractions print(f'\nTotal : {fractions.Fraction(str(round(p0,5)))}') ''' 0 0 0 : 0.6 * 0.5 * 0.3 * 0.0 = 0.0000 0 0 1 : 0.6 * 0.5 * 0.7 * 0.2 = 0.0420 0 1 0 : 0.6 * 0.5 * 0.3 * 0.2 = 0.0180 0 1 1 : 0.6 * 0.5 * 0.7 * 0.6 = 0.1260 1 0 0 : 0.4 * 0.5 * 0.3 * 0.2 = 0.0120 1 0 1 : 0.4 * 0.5 * 0.7 * 0.6 = 0.0840 1 1 0 : 0.4 * 0.5 * 0.3 * 0.6 = 0.0360 1 1 1 : 0.4 * 0.5 * 0.7 * 1.0 = 0.1400 Total : 229/500 '''
实用模块之类方法函数
小数转分数（以下基本是为了凑字数，不喜勿喷忽略即可）
fractions.Fraction
| Examples | -------- | | >>> Fraction(10, -8) | Fraction(-5, 4) | >>> Fraction(Fraction(1, 7), 5) | Fraction(1, 35) | >>> Fraction(Fraction(1, 7), Fraction(2, 3)) | Fraction(3, 14) | >>> Fraction('314') | Fraction(314, 1) | >>> Fraction('-35/4') | Fraction(-35, 4) | >>> Fraction('3.1415') # conversion from numeric string | Fraction(6283, 2000) | >>> Fraction('-47e-2') # string may include a decimal exponent | Fraction(-47, 100) | >>> Fraction(1.47) # direct construction from float (exact conversion) | Fraction(6620291452234629, 4503599627370496) | >>> Fraction(2.25) | Fraction(9, 4) | >>> Fraction(Decimal('1.47')) | Fraction(147, 100) | >>> Fraction('8.125') | Fraction(65, 8) | >>> print(Fraction('8.125')) | 65/8 | >>> print(Fraction(0.125)) | 1/8
另外解决概率题经常要用到排列、组合函数：
itertools.combinations
Help on class combinations in module itertools: class combinations(builtins.object) | combinations(iterable, r) | | Return successive r-length combinations of elements in the iterable. | | combinations(range(4), 3) --> (0,1,2), (0,1,3), (0,2,3), (1,2,3) | | Methods defined here: | | __getattribute__(self, name, /) | Return getattr(self, name). | | __iter__(self, /) | Implement iter(self). | | __next__(self, /) | Implement next(self). | | __reduce__(...) | Return state information for pickling. | | __setstate__(...) | Set state information for unpickling. | | __sizeof__(...) | Returns size in memory, in bytes. | | ---------------------------------------------------------------------- | Static methods defined here: | | __new__(*args, **kwargs) from builtins.type | Create and return a new object. See help(type) for accurate signature.
itertools.permutations
Help on class permutations in module itertools:
class permutations(builtins.object) | permutations(iterable, r=None) | | Return successive r-length permutations of elements in the iterable. | | permutations(range(3), 2) --> (0,1), (0,2), (1,0), (1,2), (2,0), (2,1) | | Methods defined here: | | __getattribute__(self, name, /) | Return getattr(self, name). | | __iter__(self, /) | Implement iter(self). | | __next__(self, /) | Implement next(self). | | __reduce__(...) | Return state information for pickling. | | __setstate__(...) | Set state information for unpickling. | | __sizeof__(...) | Returns size in memory, in bytes. | | ---------------------------------------------------------------------- | Static methods defined here: | | __new__(*args, **kwargs) from builtins.type | Create and return a new object. See help(type) for accurate signature.
还有一个可以取重复值的组合公式，知道的比较少：
tertools.combinations_with_replacement
Help on class combinations_with_replacement in module itertools:
class combinations_with_replacement(builtins.object) | combinations_with_replacement(iterable, r) | | Return successive r-length combinations of elements in the iterable allowing individual elements to have successive repeats. | | combinations_with_replacement('ABC', 2) --> AA AB AC BB BC CC" | | Methods defined here: | | __getattribute__(self, name, /) | Return getattr(self, name). | | __iter__(self, /) | Implement iter(self). | | __next__(self, /) | Implement next(self). | | __reduce__(...) | Return state information for pickling. | | __setstate__(...) | Set state information for unpickling. | | __sizeof__(...) | Returns size in memory, in bytes. | | ---------------------------------------------------------------------- | Static methods defined here: | | __new__(*args, **kwargs) from builtins.type | Create and return a new object. See help(type) for accurate signature.
