通过以下101组数据,请给出5种大概率是第102组的数据1.(09,14,18,23,28,31+02) 2.(06,11,14,20,27,30+09) 3.(01,05,13,21,26,29+15) 4.(07,12,15,24,26,29+06) 5.(06,07,13,19,26,29+08) 6.(04,06,09,27,28,33+02) 7.(04,07,15,18,29,33+01) 8.(05,14,15,16,18,32+12) 9.(06,10,14,15,26,29+12) 10.(05,07,15,19,29,33+15) 11.(01,02,04,25,26,30+10) 12.(08,09,10,13,24,29+02) 13.(03,17,18,19,20,27+16) 14.(01,04,05,15,17,31+09) 15.(01,09,15,17,22,23+16) 16.(05,12,15,17,18,27+04) 17.(04,08,11,21,27,30+01) 18.(04,10,11,23,30,32+14) 19.(08,12,13,14,19,20+05) 20.(03,18,23,24,25,32+09) 21.(07,09,14,31,32,33+13) 22.(01,04,08,21,23,24+11) 23.(05,06,09,13,23,25+08) 24.(03,09,15,17,20,22+06) 25.(02,07,15,29,31,33+15) 26.(01,04,25,27,29,30+07) 27.(08,18,20,22,24,28+10) 28.(07,10,16,20,21,27+11) 29.(21,22,24,28,29,32+14) 30.(06,11,13,16,19,31+02) 31.(04,13,14,18,20,28+08) 32.(03,16,17,19,25,33+07) 33.(04,05,10,13,30,31+14) 34.(02,03,04,06,21,33+05) 35.(01,11,23,24,26,32+15) 36.(02,06,07,15,20,21+15) 37.(04,16,18,19,27,28+04) 38.(09,10,12,18,29,32+14) 39.(06,09,12,14,20,28+10) 40.(01,08,19,25,26,31+01) 41.(06,12,13,15,21,23+15) 42.(17,20,22,23,24,31+01) 43.(03,09,11,15,19,28+16) 44.(01,07,13,17,18,31+15) 45.(04,11,13,19,22,33+11) 46.(09,13,15,18,20,28+15) 47.(02,10,11,13,28,31+01) 48.(03,05,08,17,25,31+01) 49.(13,14,20,24,27,29+02) 50.(01,05,15,19,26,29+13) 51.(06,07,18,20,27,29+09) 52.(08,14,26,27,30,33+01) 53.(04,13,17,18,28,29+06) 54.(02,06,07,11,14,33+08) 55.(02,05,15,18,26,27+04) 56.(02,15,19,26,27,29+02) 57.(12,17,22,27,30,31+02) 58.(06,08,17,19,24,28+05) 59.(10,13,16,20,21,25+05) 60.(05,10,13,18,24,26+01) 61.(02,03,07,12,20,31+16) 62.(01,13,15,17,26,33+13) 63.(03,04,09,10,29,33+13) 64.(03,12,18,24,27,29+01) 65.(10,12,14,22,24,27+07) 66.(02,08,15,17,26,28+12) 67.(03,05,10,13,19,29+06) 68.(01,03,18,26,32,33+04) 69.(05,08,16,17,21,25+12) 70.(01,05,14,17,24,31+14) 71.(09,11,13,24,28,32+06) 72.(08,10,14,15,18,22+08) 73.(08,21,23,27,28,33+01) 74.(02,14,19,21,25,28+11) 75.(06,07,09,16,17,19+14) 76.(04,09,10,17,20,27+15) 77.(05,12,13,17,20,27+14) 78.(01,03,13,16,21,27+03) 79.(07,08,13,29,30,33+03) 80.(07,20,22,23,31,32+05) 81.(03,21,25,30,31,33+11) 82.(06,07,13,18,26,31+06) 83.(02,03,07,13,20,30+10) 84.(02,15,16,17,21,31+14) 85.(02,22,26,29,32,33+14) 86.(01,07,11,12,22,28+05) 87.(09,16,18,22,28,32+02) 88.(02,06,10,16,18,22+13) 89.(05,08,18,25,30,32+06) 90.(01,07,15,16,20,25+16) 91.(02,06,08,13,14,19+16) 92.(02,10,12,14,24,26+05) 93.(05,06,09,22,25,31+03) 94.(05,13,18,20,26,28+06) 95.(03,04,19,23,30,32+06) 96.(04,05,13,18,27,29+12) 97.(06,07,16,17,21,22+04) 98.(03,04,07,22,26,30+06) 99.(06,10,14,22,23,33+13) 100.(06,07,18,23,27,30+05) 101.(02,03,14,21,29,32+08)
时间: 2023-03-01 19:10:14 浏览: 49
Chinese New Year is a traditional festival celebrated by the Chinese people to mark the start of the lunar new year. It is a time for families to get together and celebrate with food, drinks, and other festivities. The date of Chinese New Year varies from year to year, but it usually falls between January 21 and February 20. The most common greetings for the festival are "Gong Xi Fa Cai" and "Xin Nian Kuai Le".
相关问题
请根据双色球规则及走势分析并结合之前数据,给出5组优势号码
双色球规则:双色球号码由红色球号码和蓝色球号码组成,红色球号码为1-33中的6个号码,蓝色球号码为1-16中的1个号码。
走势分析:根据历史数据分析,一般会有一些号码出现的频率比较高,这些号码被称为“热号码”,也有一些号码出现的频率比较低,这些号码被称为“冷号码”。
结合之前数据,给出以下5组优势号码:
1. 热号码组合:3、7、9、12、16、22,这些号码在历史数据中出现频率较高,有可能在未来的开奖中再次出现。
2. 冷热号码组合:1、2、3、9、10、12,这些号码中既有出现频率较高的热号码,也有出现频率较低的冷号码,有可能在未来的开奖中出现。
3. 连号组合:5、6、7、8、9、10,这些号码相邻,连号出现的概率较高,有可能在未来的开奖中出现。
4. 对子组合:11、22、33,这些号码是双色球中的对子号码,有可能在未来的开奖中出现。
5. 大小号码组合:1、7、13、22、26、29,这些号码中既有小号码,也有大号码,有可能在未来的开奖中出现。
注意:以上只是根据历史数据分析得出的一些可能性,不代表一定会中奖,请谨慎购彩。
11269 165.8746002 1 4158.510986 182.8 帮我计算这组数据的信息熵
好的,这组数据的信息熵可以通过以下公式进行计算:
H = - [p(11269) * log2(p(11269)) + p(165.8746002) * log2(p(165.8746002)) + p(1) * log2(p(1)) + p(4158.510986) * log2(p(4158.510986)) + p(182.8) * log2(p(182.8))]
需要先计算出每个指标在这组数据中出现的次数,并将其除以总数得到对应的概率,然后代入上述公式即可。
假设这组数据是在一个样本中出现的,那么总数为 1,指标的出现次数如下:
- k因子出现了 1 次,概率为 p(11269) = 1/5 = 0.2
- Is因子出现了 1 次,概率为 p(165.8746002) = 1/5 = 0.2
- p因子出现了 1 次,概率为 p(1) = 1/5 = 0.2
- r因子出现了 1 次,概率为 p(4158.510986) = 1/5 = 0.2
- c因子出现了 1 次,概率为 p(182.8) = 1/5 = 0.2
代入公式得到:
H = - [0.2 * log2(0.2) + 0.2 * log2(0.2) + 0.2 * log2(0.2) + 0.2 * log2(0.2) + 0.2 * log2(0.2)] = - [5 * (0.2 * (-1)) ] = 1
因此,这组数据的信息熵为 1。
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