2021年高考數(shù)學(xué)真題和模擬題分類匯編專題14概率與統(tǒng)計(jì)【含答案】

專題14概率與統(tǒng)計(jì)/r/n一、選擇題部分/r/n1/r/n./r/n(2021?新高考全國(guó)Ⅰ卷/r/n?/r/nT8)/r/n有/r/n6/r/n個(gè)相同的球，分別標(biāo)有數(shù)字/r/n1/r/n，/r/n2/r/n，/r/n3/r/n，/r/n4/r/n，/r/n5/r/n，/r/n6/r/n，從中有放回的隨機(jī)取兩次，每次取/r/n1/r/n個(gè)球，甲表示事件/r/n“/r/n第一次取出的球的數(shù)字是/r/n1”/r/n，乙表示事件/r/n“/r/n第二次取出的球的數(shù)字是/r/n2”/r/n，丙表示事件/r/n“/r/n兩次取出的球的數(shù)字之和是/r/n8”/r/n，丁表示事件/r/n“/r/n兩次取出的球的數(shù)字之和是/r/n7”/r/n，則（）/r/nA./r/n甲與丙相互獨(dú)立/r/n B./r/n甲與丁相互獨(dú)立/r/nC./r/n乙與丙相互獨(dú)立/r/n D./r/n丙與丁相互獨(dú)立/r/nB/r/n．/r/n，/r/n故選/r/nB/r/n．/r/n2.(2021?新高考全國(guó)Ⅰ卷?T9)/r/n有一組樣本數(shù)據(jù)/r/n，/r/n，/r/n…/r/n，/r/n，由這組數(shù)據(jù)得到新樣本數(shù)據(jù)/r/n，/r/n，/r/n…/r/n，/r/n，其中/r/n(/r/n為/r/n非零常數(shù)，則（）/r/nA.兩組樣本數(shù)據(jù)的樣本平均數(shù)相同/r/nB.兩組樣本數(shù)據(jù)/r/n樣本中位數(shù)相同/r/nC.兩組樣本數(shù)據(jù)的樣本標(biāo)準(zhǔn)差相同/r/nD.兩組樣數(shù)據(jù)的樣本極差相同/r/nCD/r/n．/r/n，故方差相同，C正確；由極差的定義知：若第一組的極差為/r/n，則第二組的極差為/r/n，故極差相同，/r/nD/r/n正確；故選/r/nCD/r/n．/r/n3/r/n./r/n(2021?高考全國(guó)甲卷/r/n?理/r/nT2)/r/n為了解某地農(nóng)村經(jīng)濟(jì)情況，對(duì)該地農(nóng)戶家庭年收入進(jìn)行抽樣調(diào)查，將農(nóng)戶家庭年收入的調(diào)查數(shù)據(jù)整理得到如下頻率分布直方圖：/r/n根據(jù)此頻率分布直方圖，下面結(jié)論中不正確的是（）/r/nA./r/n該地農(nóng)戶家庭年收入低于/r/n4.5/r/n萬(wàn)元的農(nóng)戶比率估計(jì)為/r/n6%/r/nB./r/n該地農(nóng)戶家庭年收入不低于/r/n10.5/r/n萬(wàn)元的農(nóng)戶比率估計(jì)為/r/n10%/r/nC./r/n估計(jì)該地農(nóng)戶家庭年收入的平均值不超過(guò)/r/n6.5/r/n萬(wàn)元/r/nD./r/n估計(jì)該地有一半以上的農(nóng)戶，其家庭年收入介于/r/n4.5/r/n萬(wàn)元至/r/n8.5/r/n萬(wàn)元之間/r/nC/r/n．/r/n因?yàn)轭l率直方圖中的組距為/r/n1/r/n，所以各組的直方圖的高度等于頻率/r/n./r/n樣本頻率直方圖中的頻率即可作為總體的相應(yīng)比率的估計(jì)值/r/n./r/n該地農(nóng)戶家庭年收入低于/r/n4.5/r/n萬(wàn)元/r/n農(nóng)戶的比率估計(jì)值為/r/n,/r/n故/r/nA/r/n正確；該地農(nóng)戶家庭年收入不低于/r/n10.5/r/n萬(wàn)元的農(nóng)戶比率估計(jì)值為/r/n,/r/n故/r/nB/r/n正確；該地農(nóng)戶家庭年收入介于/r/n4.5/r/n萬(wàn)元至/r/n8.5/r/n萬(wàn)元之間的比例估計(jì)值為/r/n,/r/n故/r/nD/r/n正確；該地農(nóng)戶家庭年收入的平均值的估計(jì)值為/r/n(/r/n萬(wàn)元/r/n)/r/n，超過(guò)/r/n6.5/r/n萬(wàn)元，故/r/nC/r/n錯(cuò)誤/r/n./r/n綜上，給出結(jié)論中不正確的是/r/nC./r/n故選：/r/nC./r/n4.(2021?高考全國(guó)甲卷?理T10)/r/n將/r/n4/r/n個(gè)/r/n1/r/n和/r/n2/r/n個(gè)/r/n0/r/n隨機(jī)排成一行，則/r/n2/r/n個(gè)/r/n0/r/n不相鄰的概率為（）/r/nA./r/n B./r/n C./r/n D./r/nC/r/n．/r/n采用插空法，/r/n4/r/n個(gè)/r/n1/r/n產(chǎn)生/r/n5/r/n個(gè)空，分/r/n2/r/n個(gè)/r/n0/r/n相鄰和/r/n2/r/n個(gè)/r/n0/r/n不相鄰進(jìn)行求解/r/n./r/n將/r/n4/r/n個(gè)/r/n1/r/n和/r/n2/r/n個(gè)/r/n0/r/n隨機(jī)排成一行，可利用插空法，/r/n4/r/n個(gè)/r/n1/r/n產(chǎn)生/r/n5/r/n個(gè)空，/r/n若/r/n2/r/n個(gè)/r/n0/r/n相鄰，則有/r/n種排法，若/r/n2/r/n個(gè)/r/n0/r/n不相鄰，則有/r/n種排法，/r/n所以/r/n2/r/n個(gè)/r/n0/r/n不相鄰的概率為/r/n./r/n故選/r/nC./r/n5.(2021?高考全國(guó)乙卷?文T7)/r/n在區(qū)間/r/n隨機(jī)取/r/n1/r/n個(gè)數(shù)，則取到的數(shù)小于/r/n的概率為（）/r/nA./r/n B./r/n C./r/n D./r/nB/r/n．/r/n設(shè)/r/n“區(qū)間/r/n隨機(jī)取/r/n1/r/n個(gè)數(shù)”/r/n，/r/n“取到的數(shù)小于/r/n”/r/n，所以/r/n．/r/n故選：/r/nB/r/n．/r/n6.(2021?江蘇鹽城三模?T/r/n9/r/n)/r/n已知/r/nX/r/n～/r/nN/r/n(/r/nμ/r/n1/r/n，/r/nσ/r/n1/r/n2/r/n)/r/n，/r/nY/r/n～/r/nN/r/n(/r/nμ/r/n2/r/n，/r/nσ/r/n2/r/n2/r/n)/r/n，/r/nμ/r/n1/r/n＞/r/nμ/r/n2/r/n，/r/nσ/r/n1/r/n＞/r/n0/r/n，/r/nσ/r/n2/r/n＞/r/n0/r/n，則下列結(jié)論中一定成立的有/r/nA/r/n．若/r/nσ/r/n1/r/n＞/r/nσ/r/n2/r/n，則/r/nP/r/n(|/r/nX/r/n－/r/nμ/r/n1/r/n|/r/n≤/r/n1)/r/n＜/r/nP/r/n(|/r/nY/r/n－/r/nμ/r/n2/r/n|/r/n≤/r/n1)/r/nB/r/n．若/r/nσ/r/n1/r/n＞/r/nσ/r/n2/r/n，則/r/nP/r/n(|/r/nX/r/n－/r/nμ/r/n1/r/n|/r/n≤/r/n1)/r/n＞/r/nP/r/n(|/r/nY/r/n－/r/nμ/r/n2/r/n|/r/n≤/r/n1)/r/nC/r/n．若/r/nσ/r/n1/r/n＝/r/nσ/r/n2/r/n，則/r/nP/r/n(/r/nX/r/n＞/r/nμ/r/n2/r/n)/r/n＋/r/nP/r/n(/r/nY/r/n＞/r/nμ/r/n1/r/n)/r/n＝/r/n1/r/nD/r/n．若/r/nσ/r/n1/r/n＝/r/nσ/r/n2/r/n，則/r/nP/r/n(/r/nX/r/n＞/r/nμ/r/n2/r/n)/r/n＋/r/nP/r/n(/r/nY/r/n＞/r/nμ/r/n1/r/n)/r/n＜/r/n1/r/nAC/r/n．/r/n【考點(diǎn)】正態(tài)分布的應(yīng)用/r/n法一：由題意可知，對(duì)于選項(xiàng)/r/nAB/r/n，若/r/nσ/r/n1/r/n＞/r/nσ/r/n2/r/n，則/r/nY/r/n分布更加集中，則在相同區(qū)間范圍/r/nY/r/n的相對(duì)概率更大，所以/r/nP/r/n(|/r/nX/r/n－/r/nμ/r/n1/r/n|/r/n≤/r/n1)/r/n＜/r/nP/r/n(|/r/nY/r/n－/r/nμ/r/n2/r/n|/r/n≤/r/n1)/r/n，所以選項(xiàng)/r/nA/r/n正確，選項(xiàng)/r/nB/r/n錯(cuò)誤；對(duì)于選項(xiàng)/r/nCD/r/n，由正態(tài)分布的性質(zhì)可得，/r/nP/r/n(/r/nY/r/n＞/r/nμ/r/n1/r/n)/r/n＝/r/nP/r/n(/r/nX/r/n≤/r/nμ/r/n2/r/n)/r/n，又/r/nP/r/n(/r/nX/r/n≤/r/nμ/r/n2/r/n)/r/n＋/r/nP/r/n(/r/nX/r/n＞/r/nμ/r/n2/r/n)/r/n＝/r/n1/r/n，所以/r/nP/r/n(/r/nX/r/n＞/r/nμ/r/n2/r/n)/r/n＋/r/nP/r/n(/r/nY/r/n＞/r/nμ/r/n1/r/n)/r/n＝/r/n1/r/n，所以選項(xiàng)/r/nC/r/n正確，選項(xiàng)/r/nD/r/n錯(cuò)誤；綜上，答案選/r/nAC/r/n．/r/n法二：由題意可知，可把正態(tài)分布標(biāo)準(zhǔn)化，即/r/nEQ/r/n\/r/nF/r/n(/r/nX/r/n－/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(1),/r/nσ/r/n\/r/nS/r/n\/r/nDO/r/n(1）/r/n＝/r/nZ/r/n＝/r/nEQ/r/n\/r/nF/r/n(/r/nY/r/n－/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(2),/r/nσ/r/n\/r/nS/r/n\/r/nDO/r/n(2）/r/n，則/r/nZ/r/n～/r/nN/r/n(0/r/n，/r/n1)/r/n，對(duì)于選項(xiàng)/r/nAB/r/n，若/r/nσ/r/n1/r/n＞/r/nσ/r/n2/r/n，則/r/nP/r/n(|/r/nX/r/n－/r/nμ/r/n1/r/n|/r/n≤/r/n1)/r/n＝/r/nP/r/n(|/r/nZ/r/n|/r/n≤/r/nEQ/r/n\/r/nF/r/n(1,/r/nσ/r/n\/r/nS/r/n\/r/nDO/r/n(1）/r/n)/r/n，/r/nP/r/n(|/r/nY/r/n－/r/nμ/r/n2/r/n|/r/n≤/r/n1)/r/n＝/r/nP/r/n(|/r/nZ/r/n|/r/n≤/r/nEQ/r/n\/r/nF/r/n(1,/r/nσ/r/n\/r/nS/r/n\/r/nDO/r/n(2）/r/n)/r/n，因?yàn)?r/nσ/r/n1/r/n＞/r/nσ/r/n2/r/n＞/r/n0/r/n，所以/r/nEQ/r/n\/r/nF/r/n(1,/r/nσ/r/n\/r/nS/r/n\/r/nDO/r/n(1）/r/n＜/r/nEQ/r/n\/r/nF/r/n(1,/r/nσ/r/n\/r/nS/r/n\/r/nDO/r/n(2）/r/n，所以/r/nP/r/n(|/r/nX/r/n－/r/nμ/r/n1/r/n|/r/n≤/r/n1)/r/n＜/r/nP/r/n(|/r/nY/r/n－/r/nμ/r/n2/r/n|/r/n≤/r/n1)/r/n，所以選項(xiàng)/r/nA/r/n正確，選項(xiàng)/r/nB/r/n錯(cuò)誤；對(duì)于選項(xiàng)/r/nCD/r/n，若/r/nσ/r/n1/r/n＝/r/nσ/r/n2/r/n，則/r/nP/r/n(/r/nX/r/n＞/r/nμ/r/n2/r/n)/r/n＝/r/nP/r/n(/r/nZ/r/n＞/r/nEQ/r/n\/r/nF/r/n(/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(2)/r/n－/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(1),/r/nσ/r/n)/r/n)/r/n，/r/nP/r/n(/r/nY/r/n＞/r/nμ/r/n1/r/n)/r/n＝/r/nP/r/n(/r/nZ/r/n＞/r/nEQ/r/n\/r/nF/r/n(/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(1)/r/n－/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(2),/r/nσ/r/n)/r/n)/r/n，所以/r/nP/r/n(/r/nX/r/n＞/r/nμ/r/n2/r/n)/r/n＋/r/nP/r/n(/r/nY/r/n＞/r/nμ/r/n1/r/n)/r/n＝/r/nP/r/n(/r/nZ/r/n＞/r/nEQ/r/n\/r/nF/r/n(/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(2)/r/n－/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(1),/r/nσ/r/n)/r/n)/r/n＋/r/nP/r/n(/r/nZ/r/n≤/r/nEQ/r/n\/r/nF/r/n(/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(1)/r/n－/r/nμ/r/n\/r/nS/r/n\/r/nDO/r/n(2),/r/nσ/r/n)/r/n)/r/n＝/r/n1/r/n，所以選項(xiàng)/r/nC/r/n正確，選項(xiàng)/r/nD/r/n錯(cuò)誤；綜上，答案選/r/nAC/r/n．/r/n7.(2021?河南開(kāi)封三模?文理T/r/n4/r/n)/r/n2/r/n021/r/n年開(kāi)始，我省將試行“/r/n3+1+2/r/n“的普通高考新模式，即除語(yǔ)文、數(shù)學(xué)、外語(yǔ)/r/n3/r/n門(mén)必選科目外，考生再?gòu)奈锢怼v史中選/r/n1/r/n門(mén)，從化學(xué)、生物、地理、政治中選/r/n2/r/n門(mén)作為選考科目．為了幫助學(xué)生合理選科，某中學(xué)將高一每個(gè)學(xué)生的六門(mén)科目綜合成績(jī)按比例均縮放成/r/n5/r/n分制，繪制成雷達(dá)圖．甲同學(xué)的成績(jī)雷達(dá)圖如圖所示，下面敘述一定不正確的是（）/r/nA/r/n．甲的物理成績(jī)領(lǐng)先年級(jí)平均分最多/r/n /r/nB/r/n．甲有/r/n2/r/n個(gè)科目的成績(jī)低于年級(jí)平均分/r/n /r/nC/r/n．甲的成績(jī)從高到低的前/r/n3/r/n個(gè)科目依次是地理、化學(xué)、歷史/r/n /r/nD/r/n．對(duì)甲而言，物理、化學(xué)、地理是比較理想的一種選科結(jié)果/r/nC/r/n．/r/n甲的成績(jī)從高到低的前/r/n3/r/n個(gè)科目依次是地理、化學(xué)、生物（物理），/r/nC/r/n選項(xiàng)錯(cuò)/r/n．/r/n8.(2021?河南開(kāi)封三模?文T/r/n10/r/n．)三/r/n人制足球（也稱為籠式足球）以其獨(dú)特的魅力，吸引著中國(guó)眾多的業(yè)余足球愛(ài)好者．在某次三人制足球傳球訓(xùn)練中，/r/nA/r/n隊(duì)有甲、乙、丙三名隊(duì)員參加，甲、乙、丙三人都等可能地將球傳給另外兩位隊(duì)友中的一個(gè)人．若由甲開(kāi)始發(fā)球（記為第一次傳球），則第四次仍由甲傳球的概率是（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nA/r/n．/r/n所有傳球方法共有：/r/n甲→乙→甲→乙；甲→乙→甲→丙；甲→乙→丙→甲；甲→乙→丙→乙；/r/n甲→丙→甲→乙；甲→丙→甲→丙；甲→丙→乙→甲；甲→丙→乙→丙．/r/n則共有/r/n8/r/n種方法．第四次仍由甲傳球有/r/n2/r/n情況，/r/n∴第四次仍由甲傳球的概率/r/nP/r/n＝/r/n＝/r/n．/r/n9.(2021?安徽宿州三模?文T/r/n3/r/n．)教/r/n育部辦公廳于/r/n2021/r/n年/r/n1/r/n月/r/n18/r/n日發(fā)布了《關(guān)于加強(qiáng)中小學(xué)生手機(jī)管理工作的通知》，通知要求中小學(xué)生原則上不得將個(gè)人手機(jī)帶入校園．某學(xué)校為了解/r/n2000/r/n名學(xué)生的手機(jī)使用情況，將這些學(xué)生編號(hào)為/r/n1/r/n，/r/n2/r/n，/r/n..../r/n，/r/n2000/r/n，從這些學(xué)生中用系統(tǒng)抽樣方法抽取/r/n200/r/n名學(xué)生進(jìn)行調(diào)查．若/r/n58/r/n號(hào)學(xué)生被抽到，則下面/r/n4/r/n名學(xué)生中被抽到的是（）/r/nA/r/n．/r/n9/r/n號(hào)學(xué)生/r/n /r/nB/r/n．/r/n300/r/n號(hào)學(xué)生/r/n /r/nC/r/n．/r/n618/r/n號(hào)學(xué)生/r/n /r/nD/r/n．/r/n816/r/n號(hào)學(xué)生/r/nC/r/n．/r/n記被抽取到的學(xué)生的編號(hào)為/r/n{/r/na/r/nn/r/n}/r/n，則/r/n{/r/na/r/nn/r/n}/r/n為等差數(shù)列，公差為/r/nd/r/n＝/r/n＝/r/n10/r/n，/r/n所以/r/na/r/nn/r/n＝/r/na/r/n1/r/n+10/r/n（/r/nn/r/n﹣/r/n1/r/n），由/r/na/r/nn/r/n＝/r/n58/r/n，解得/r/na/r/n1/r/n＝/r/n8/r/n，所以/r/na/r/nn/r/n＝/r/n10/r/nn/r/n﹣/r/n2/r/n，/r/n所以編號(hào)為/r/n618/r/n的學(xué)生可以被抽取到．/r/n10.(2021?安徽宿州三模?文T4理T/r/n3/r/n．)/r/n我國(guó)古代著名數(shù)學(xué)家祖沖之早在/r/n1500/r/n多年前就算出圓周率/r/nπ/r/n的近似值在/r/n3/r/n.和/r/n3/r/n.之/r/n間，這是我國(guó)古代數(shù)學(xué)的一大成就．我們知道用均勻投點(diǎn)的模擬方法，也可以獲得問(wèn)題的近似解．如圖，一個(gè)圓內(nèi)切于一個(gè)正方形，現(xiàn)利用模擬方法向正方形內(nèi)均勻投點(diǎn)，若投點(diǎn)落在圓內(nèi)的概率為/r/n，則估計(jì)圓周率的值為（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/n由幾何概型得：/r/nP/r/n＝/r/n＝/r/n，∴/r/nπ/r/n＝/r/n．/r/nA/r/n．/r/n11.(2021?江西上饒三模?理T/r/n3/r/n．)已/r/n知隨機(jī)變量/r/nξ/r/n服從正態(tài)分布/r/nN/r/n（/r/n3/r/n，σ/r/n2/r/n），/r/nP/r/n（/r/nξ/r/n≤/r/n6/r/n）＝/r/n0.84/r/n，則/r/nP/r/n（/r/nξ/r/n≤/r/n0/r/n）＝（）/r/nA/r/n．/r/n0.16/r/n /r/nB/r/n．/r/n0.34/r/n /r/nC/r/n．/r/n0.66/r/n /r/nD/r/n．/r/n0.84/r/nA/r/n．/r/n∵/r/nP/r/n（/r/nξ/r/n＞/r/n6/r/n）＝/r/n1/r/n﹣/r/n0.84/r/n＝/r/n0.16/r/n，∴/r/nP/r/n（/r/nξ/r/n≤/r/n0/r/n）＝/r/nP/r/n（/r/nξ/r/n＞/r/n6/r/n）＝/r/n0.16/r/n．/r/n12.(2021?山東聊城三模?T/r/n9./r/n)/r/n對(duì)具有相關(guān)關(guān)系的兩個(gè)變量/r/nx/r/n和/r/ny/r/n進(jìn)行回歸分折時(shí)，經(jīng)過(guò)隨機(jī)抽樣獲得成對(duì)的樣本點(diǎn)數(shù)據(jù)/r/n(/r/nx/r/n1/r/nA./r/n若兩變量/r/nx/r/n，/r/ny/r/n具有線性相關(guān)關(guān)系，則回歸直線至少經(jīng)過(guò)一個(gè)樣本點(diǎn)/r/n

/r/nB./r/n若兩變量/r/nx/r/n，/r/ny/r/n具有線性相關(guān)關(guān)系，則回歸直線一定經(jīng)過(guò)樣本點(diǎn)中心/r/n(/r/nx/r/n,/r/ny/r/n)/r/n

/r/nC./r/n若以模型/r/ny=a/r/ne/r/nbx/r/n擬合該組數(shù)據(jù)，為了求出回歸方程，設(shè)/r/nz=/r/nln/r/ny/r/n，將其變換后得到線性方程/r/nz=6x+/r/nln/r/n3/r/n，則/r/na/r/n，/r/nb/r/n的估計(jì)值分別是/r/n3/r/n和/r/nB,C,D/r/n．/r/n【考點(diǎn)】線性回歸方程，可線性化的回歸分析/r/n若兩變量/r/nx/r/n，/r/ny/r/n具有線性相關(guān)關(guān)系，即滿足/r/ny=/r/nb/r/nx+/r/na/r/n，則一定滿足/r/ny/r/n=/r/nb/r/n若以模型/r/ny=a/r/ne/r/nh/r/nx/r/n擬合該組數(shù)據(jù)，/r/nz=/r/nln/r/ny=bx+/r/nln/r/na=6x+/r/nln/r/n3/r/n，故/r/na=3,b=6/r/n，/r/nC/r/n符合題意；用/r/nR/r/n【分析】根據(jù)線性相關(guān)關(guān)系可判斷/r/nA/r/n錯(cuò)誤，/r/nB/r/n正確。根據(jù)擬合曲線關(guān)系可判斷/r/nC/r/n正確，/r/nD/r/n正確。/r/n13.(2021?山東聊城三模?T/r/n6./r/n)在/r/n某次脫貧攻堅(jiān)表彰會(huì)上，共有/r/n36/r/n人受到表彰，其中男性多于女性，現(xiàn)從中隨機(jī)選出/r/n2/r/n人作為代表上臺(tái)領(lǐng)獎(jiǎng)，若選出的兩人性別相同的概率為/r/n1/r/n2/r/n，則受表彰人員中男性人數(shù)為（）/r/nA.15B.18/r/nC.21/r/nD.15/r/n或/r/n21/r/nC/r/n．/r/n【考點(diǎn)】古典概型及其概率計(jì)算公式，組合及組合數(shù)公式，一元二次方程/r/n設(shè)男性有/r/nx/r/n人，則女性有/r/n36/r/n-/r/nx/r/n∵/r/n男性多于女性，/r/n∴/r/nx>36/r/n-/r/nx/r/n，即/r/n∵/r/n選出的兩人性別相同的概率為/r/n1/r/n2/r/n．/r/n∴/r/nC/r/nx/r/n2/r/n∴/r/nx=21/r/n或/r/nx=15/r/n（舍）/r/n．/r/n所以男性有/r/n21/r/n人/r/n．/r/n故/r/nC./r/n【分析】根據(jù)古典概率可得/r/nC/r/nx/r/n2/r/n+/r/nC/r/n14.(2021?四川內(nèi)江三模?理T/r/n4/r/n．)為/r/n了普及環(huán)保知識(shí)，增強(qiáng)環(huán)保意識(shí)，某大學(xué)隨機(jī)抽取/r/n30/r/n名學(xué)生參加環(huán)保知識(shí)測(cè)試（十分制）如圖所示，假設(shè)得分值的中位數(shù)為/r/nm/r/ne/r/n，眾數(shù)為/r/nm/r/no/r/n，平均值為/r/n，則（）/r/nA/r/n．/r/nm/r/ne/r/n＝/r/nm/r/no/r/n＝/r/n /r/nB/r/n．/r/nm/r/ne/r/n＝/r/nm/r/no/r/n＜/r/n /r/nC/r/n．/r/nm/r/ne/r/n＜/r/nm/r/no/r/n＜/r/n /r/nD/r/n．/r/nm/r/no/r/n＜/r/nm/r/ne/r/n＜/r/nD/r/n．/r/n由圖知/r/nm/r/n0/r/n＝/r/n5/r/n，/r/n有中位數(shù)的定義應(yīng)該是第/r/n15/r/n個(gè)數(shù)與第/r/n16/r/n個(gè)數(shù)的平均值，/r/n由圖知將數(shù)據(jù)從大到小排第/r/n15/r/n個(gè)數(shù)是/r/n6/r/n，第/r/n16/r/n個(gè)數(shù)是/r/n6/r/n，/r/n所以/r/n＞/r/n3.9/r/n．/r/n15.(2021?重慶名校聯(lián)盟三模?T/r/n9/r/n．)/r/n空氣質(zhì)量指數(shù)大小分為五級(jí)，指數(shù)越大說(shuō)明污染的情況越嚴(yán)重，對(duì)人體危害越大．指數(shù)范圍在：/r/n[0/r/n，/r/n50]/r/n，/r/n[51/r/n，/r/n100]/r/n，/r/n[101/r/n，/r/n200]/r/n，/r/n[201/r/n，/r/n300]/r/n，/r/n[301/r/n，/r/n500]/r/n分別對(duì)應(yīng)“優(yōu)”、“良”、“輕度污染“、“中度污染”“重度污染”五個(gè)等級(jí)，下面是某市連續(xù)/r/n14/r/n天的空氣質(zhì)量指數(shù)變化趨勢(shì)圖，下列說(shuō)法中正確的是（）/r/nA/r/n．從/r/n2/r/n日到/r/n5/r/n日空氣質(zhì)量越來(lái)越好/r/n /r/nB/r/n．這/r/n14/r/n天中空氣質(zhì)量指數(shù)的極差為/r/n195/r/n /r/nC/r/n．這/r/n14/r/n天中空氣質(zhì)量指數(shù)的中位數(shù)是/r/n103.5/r/n /r/nD/r/n．這/r/n14/r/n天中空氣質(zhì)量指數(shù)為“良”的頻率為/r/nBC/r/n．/r/n對(duì)于/r/nA/r/n，由折線圖可知，從/r/n2/r/n日到/r/n5/r/n日空氣質(zhì)量指數(shù)越來(lái)越大，所以空氣質(zhì)量越來(lái)越差，故選項(xiàng)/r/nA/r/n錯(cuò)誤；/r/n對(duì)于/r/nB/r/n，這/r/n14/r/n天中空氣質(zhì)量指數(shù)的極差為/r/n220/r/n﹣/r/n25/r/n＝/r/n195/r/n，故選項(xiàng)/r/nB/r/n正確；/r/n對(duì)于/r/nC/r/n，這/r/n14/r/n天中空氣質(zhì)量指數(shù)為/r/n25/r/n，/r/n37/r/n，/r/n40/r/n，/r/n57/r/n，/r/n79/r/n，/r/n86/r/n，/r/n86/r/n，/r/n121/r/n，/r/n143/r/n，/r/n158/r/n，/r/n160/r/n，/r/n160/r/n，/r/n217/r/n，/r/n220/r/n，所以中位數(shù)是（/r/n86+121/r/n）÷/r/n2/r/n＝/r/n103.5/r/n，故選項(xiàng)/r/nC/r/n正確；/r/n對(duì)于/r/nD/r/n，這/r/n14/r/n天中空氣質(zhì)量指數(shù)為“良”的頻率為/r/n，故選項(xiàng)/r/nD/r/n錯(cuò)誤．/r/n16.(2021?重慶名校聯(lián)盟三模?T/r/n4/r/n．)孿/r/n生素?cái)?shù)猜想是希爾伯特在/r/n1900/r/n年提出的/r/n23/r/n問(wèn)題中的第/r/n8/r/n個(gè)：存在無(wú)窮多個(gè)素?cái)?shù)/r/np/r/n，使得/r/np/r/n+2/r/n是素?cái)?shù)，素?cái)?shù)對(duì)（/r/np/r/n，/r/np/r/n+2/r/n）稱為孿生素?cái)?shù)．/r/n2013/r/n年華人數(shù)學(xué)家張益唐發(fā)表的論文《素?cái)?shù)間的有界距離》第一次證明了存在無(wú)窮多組間距小于定值的素?cái)?shù)對(duì)．那么在不超過(guò)/r/n16/r/n的素?cái)?shù)中任意取出不同的兩個(gè)，可組成孿生素?cái)?shù)的概率為（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nA/r/n．/r/n不超過(guò)/r/n16/r/n的素?cái)?shù)有/r/n2/r/n，/r/n3/r/n，/r/n5/r/n，/r/n7/r/n，/r/n11/r/n，/r/n13/r/n，/r/n在不超過(guò)/r/n16/r/n的素?cái)?shù)中任意取出不同的兩個(gè)，/r/n基本事件總數(shù)/r/nn/r/n＝/r/n＝/r/n15/r/n，/r/n可組成孿生素?cái)?shù)包含的基本事件有：/r/n（/r/n3/r/n，/r/n5/r/n），（/r/n5/r/n，/r/n7/r/n），（/r/n11/r/n，/r/n13/r/n），共/r/n3/r/n個(gè)，/r/n∴在不超過(guò)/r/n16/r/n的素?cái)?shù)中任意取出不同的兩個(gè)，可組成孿生素?cái)?shù)的概率為/r/nP/r/n＝/r/n．/r/n17.(2021?安徽蚌埠三模?文T/r/n5/r/n．)/r/n國(guó)家統(tǒng)計(jì)局官方網(wǎng)站/r/n2021/r/n年/r/n2/r/n月/r/n28/r/n日發(fā)布了《中華人民共和國(guó)/r/n2020/r/n年國(guó)民經(jīng)濟(jì)和社會(huì)發(fā)展統(tǒng)計(jì)公報(bào)》，全面展示了一年來(lái)全國(guó)人民頑強(qiáng)奮斗取得的令世界矚目、可載入史冊(cè)的偉大成就．如圖是/r/n2016/r/n﹣/r/n2020/r/n年國(guó)內(nèi)生產(chǎn)總值及其增長(zhǎng)速度統(tǒng)計(jì)圖和三次產(chǎn)業(yè)增加值占國(guó)內(nèi)生產(chǎn)總值比重統(tǒng)計(jì)圖．/r/n給出下列說(shuō)法：/r/n①/r/n從/r/n2016/r/n年至/r/n2020/r/n年國(guó)內(nèi)生產(chǎn)總值逐年遞增；/r/n②/r/n從/r/n2016/r/n年至/r/n2020/r/n年國(guó)內(nèi)生產(chǎn)總值增長(zhǎng)速度逐年遞減；/r/n③/r/n從/r/n2016/r/n年至/r/n2020/r/n年第三產(chǎn)業(yè)增加值占國(guó)內(nèi)生產(chǎn)總值比重逐年遞增；/r/n④/r/n從/r/n2016/r/n年至/r/n2020/r/n年第二產(chǎn)業(yè)增加值占國(guó)內(nèi)生產(chǎn)總值比重逐年遞減．/r/n其中正確的是（）/r/nA/r/n．/r/n①④/r/n /r/nB/r/n．/r/n②③/r/n /r/nC/r/n．/r/n②④/r/n /r/nD/r/n．/r/n①③/r/nD/r/n．/r/n對(duì)于/r/n①/r/n，由圖/r/n1/r/n可知，從/r/n2016/r/n年到/r/n2020/r/n年國(guó)內(nèi)生產(chǎn)總值數(shù)不斷的增大，/r/n條形圖中對(duì)應(yīng)的長(zhǎng)方形的高度不斷升高，故選項(xiàng)/r/n①/r/n正確；/r/n對(duì)于/r/n②/r/n，由圖/r/n2/r/n可知，在/r/n2016/r/n年到/r/n2017/r/n年國(guó)內(nèi)生產(chǎn)總值增長(zhǎng)的折線是上升的，/r/n從/r/n6.8/r/n到/r/n6.9/r/n，故選項(xiàng)/r/n②/r/n錯(cuò)誤；/r/n對(duì)于/r/n③/r/n，由圖/r/n2/r/n可知，/r/n2016/r/n年到/r/n2020/r/n年第三產(chǎn)業(yè)增加值占國(guó)內(nèi)生產(chǎn)總值比重/r/n從/r/n52.4/r/n→/r/n52.7/r/n→/r/n53.3/r/n→/r/n54.3/r/n→/r/n54.5/r/n，是不斷增加的，故選項(xiàng)/r/n③/r/n正確；/r/n對(duì)應(yīng)/r/n④/r/n，由圖/r/n2/r/n可知，在/r/n2016/r/n年到/r/n2017/r/n年第二產(chǎn)業(yè)增加值/r/n占國(guó)內(nèi)生產(chǎn)總值比重由/r/n39.6/r/n上升到了/r/n39.9/r/n，故選項(xiàng)/r/n④/r/n錯(cuò)誤．/r/n18.(2021?上海嘉定三模?T/r/n10/r/n．)有/r/n大小相同的紅、黃、藍(lán)三種顏色的小球各/r/n3/r/n個(gè)，且每種顏色的/r/n3/r/n個(gè)小球上分別標(biāo)注號(hào)碼/r/n1/r/n、/r/n2/r/n、/r/n3/r/n，從中任取/r/n3/r/n個(gè)球，則取出的/r/n3/r/n個(gè)球顏色齊全但號(hào)碼不全的概率是/r/n/r/n/r/n．/r/n．/r/n反面法：取出的/r/n3/r/n個(gè)球顏色齊全但號(hào)碼齊全的情況為/r/n6/r/n種，/r/n取出的/r/n3/r/n個(gè)球顏色齊全但號(hào)碼不全的概率是/r/n．/r/n19.(2021?貴州畢節(jié)三模?文T/r/n3/r/n．)/r/n一袋中裝有除顏色外完全相同的/r/n4/r/n個(gè)白球和/r/n5/r/n個(gè)黑球，從中有放回的摸球/r/n3/r/n次，每次摸一個(gè)球．用模擬實(shí)驗(yàn)的方法，讓計(jì)算機(jī)產(chǎn)生/r/n1/r/n～/r/n9/r/n的隨機(jī)數(shù)，若/r/n1/r/n～/r/n4/r/n代表白球，/r/n5/r/n～/r/n9/r/n代表黑球，每三個(gè)為一組，產(chǎn)生如下/r/n20/r/n組隨機(jī)數(shù)：/r/n9179661/r/n9/r/n1/r/n925271/r/n93/r/n2/r/n735458569683/r/n4312573/r/n9/r/n3/r/n627556/r/n48/r/n8/r/n812184537989/r/n則三次摸出的球中恰好有兩次是白球的概率近似為（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nB/r/n．/r/n20/r/n組隨機(jī)數(shù)恰好有兩個(gè)是/r/n1/r/n，/r/n2/r/n，/r/n3/r/n，/r/n4/r/n的有/r/n191/r/n，/r/n171/r/n，/r/n932/r/n，/r/n393/r/n，/r/n812/r/n，/r/n184/r/n，共/r/n6/r/n個(gè)，因此三次摸出的球中恰好有兩次是白球的概率近似為/r/n．/r/n20.(2021?遼寧朝陽(yáng)三模?T/r/n8/r/n．)在/r/n三棱柱/r/nABC/r/n﹣/r/nA/r/n1/r/nB/r/n1/r/nC/r/n1/r/n中，/r/nD/r/n為側(cè)棱/r/nCC/r/n1/r/n的中點(diǎn)，從該三棱柱的九條棱中隨機(jī)選取兩條，則這兩條棱所在直線至少有一條與直線/r/nBD/r/n異面的概率是（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nB/r/n．/r/n∵在三棱柱/r/nABC/r/n﹣/r/nA/r/n1/r/nB/r/n1/r/nC/r/n1/r/n中，/r/nD/r/n為側(cè)棱/r/nCC/r/n1/r/n的中點(diǎn)，/r/n∴該三棱柱的九條棱中與/r/nBD/r/n異面的棱有/r/n5/r/n條，從該三棱柱的九條棱中隨機(jī)選取兩條，/r/n基本事件總數(shù)/r/nn/r/n＝/r/n＝/r/n36/r/n，這兩條棱所在直線至少有一條與直線/r/nBD/r/n異面包含的基本事件個(gè)數(shù)為：/r/nm/r/n＝/r/n+/r/n＝/r/n26/r/n，則這兩條棱所在直線至少有一條與直線/r/nBD/r/n異面的概率/r/nP/r/n＝/r/n＝/r/n＝/r/n．/r/n21.(2021?河南濟(jì)源平頂山許昌三模?文T/r/n3/r/n．)/r/n某交通廣播電臺(tái)在正常播音期間，每個(gè)整點(diǎn)都會(huì)進(jìn)行報(bào)時(shí)．某出租車(chē)司機(jī)在該交通廣播電臺(tái)正常播音期間，打開(kāi)收音機(jī)想收聽(tīng)電臺(tái)整點(diǎn)報(bào)時(shí)，則他等待時(shí)間不超過(guò)/r/n5/r/n分鐘的概率為（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nB/r/n．/r/n設(shè)電臺(tái)的整點(diǎn)報(bào)時(shí)之間某刻的時(shí)間/r/nx/r/n，由題意可得，/r/n0/r/n≤/r/nx/r/n≤/r/n60/r/n，/r/n則等待的時(shí)間不超過(guò)/r/n5/r/n分鐘的概率為/r/nP/r/n＝/r/n．/r/n22.(2021?江蘇常數(shù)三模?T/r/n2/r/n．)/r/n若隨機(jī)變量/r/nX/r/n～/r/nB/r/n（/r/n5/r/n，/r/np/r/n），/r/n，則/r/nE/r/n（/r/nX/r/n）＝（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nD/r/n．/r/n因?yàn)?r/nX/r/n～/r/nB/r/n（/r/n5/r/n，/r/np/r/n），/r/n，/r/n則/r/n，解得/r/n，所以/r/n．/r/n23.(2021?湖南三模?T/r/n8/r/n．)/r/n在一次“概率”相關(guān)的研究性活動(dòng)中，老師在每個(gè)箱子中裝了/r/n10/r/n個(gè)小球，其中/r/n9/r/n個(gè)是白球，/r/n1/r/n個(gè)是黑球，用兩種方法讓同學(xué)們來(lái)摸球．方法一：在/r/n20/r/n箱中各任意摸出一個(gè)小球；方法二：在/r/n10/r/n箱中各任意摸出兩個(gè)小球．將方法一、二至少能摸出一個(gè)黑球的概率分別記為/r/np/r/n1/r/n和/r/np/r/n2/r/n，則（）/r/nA/r/n．/r/np/r/n1/r/n＜/r/np/r/n2/r/n /r/nB/r/n．/r/np/r/n1/r/n＝/r/np/r/n2/r/n /r/nC/r/n．/r/np/r/n1/r/n＞/r/np/r/n2/r/n /r/nD/r/n．以上三種情況都有可能/r/nA/r/n．/r/n根據(jù)題意，按方法一抽取，每箱中黑球被抽取的概率為/r/n，則沒(méi)有抽到黑球的概率為/r/n1/r/n﹣/r/n＝/r/n，則至少能摸出一個(gè)黑球的概率/r/nP/r/n1/r/n＝/r/n1/r/n﹣（/r/n）/r/n20/r/n，按方法一抽取，每箱中黑球被抽取的概率為/r/n，則沒(méi)有抽到黑球的概率為/r/n1/r/n﹣/r/n＝/r/n，則至少能摸出一個(gè)黑球的概率/r/nP/r/n2/r/n＝/r/n1/r/n﹣（/r/n）/r/n10/r/n，則有/r/nP/r/n1/r/n﹣/r/nP/r/n2/r/n＝/r/n[1/r/n﹣（/r/n）/r/n20/r/n]/r/n﹣/r/n[1/r/n﹣（/r/n）/r/n10/r/n]/r/n＝（/r/n）/r/n10/r/n﹣（/r/n）/r/n20/r/n＝（/r/n）/r/n10/r/n﹣（/r/n）/r/n10/r/n＜/r/n0/r/n，故/r/nP/r/n1/r/n＜/r/nP/r/n2/r/n．/r/n24.(2021?湖南三模?T/r/n3/r/n．)/r/n每年的/r/n3/r/n月/r/n15/r/n日是“國(guó)際消費(fèi)者權(quán)益日”，某地市場(chǎng)監(jiān)管局在當(dāng)天對(duì)某市場(chǎng)的/r/n20/r/n家肉制品店、/r/n100/r/n家糧食加工品店和/r/n15/r/n家乳制品店進(jìn)行抽檢，要用分層抽樣的方法從中抽檢/r/n27/r/n家，則糧食加工品店需要被抽檢（）/r/nA/r/n．/r/n20/r/n家/r/n /r/nB/r/n．/r/n10/r/n家/r/n /r/nC/r/n．/r/n15/r/n家/r/n /r/nD/r/n．/r/n25/r/n家/r/nA/r/n．/r/n根據(jù)分層抽樣原理知，糧食加工品店需要被抽檢/r/n27/r/n×/r/n＝/r/n20/r/n（家）．/r/n25.(2021?福建寧德三模?T10)/r/n某校研究性學(xué)習(xí)小組根據(jù)某市居民人均消費(fèi)支出的統(tǒng)計(jì)數(shù)據(jù)，制作/r/n2018/r/n年人均消費(fèi)支出條形圖/r/n(/r/n單位：元/r/n)/r/n和/r/n2019/r/n年人均消費(fèi)支出餅圖/r/n(/r/n如圖/r/n)./r/n已知/r/n2019/r/n年居民人均消費(fèi)總支出比/r/n2018/r/n年居民人均消費(fèi)總支出提高/r/n8.5%/r/n，則下列結(jié)論正確的是/r/n(?)/r/n

/r/nA.2019/r/n年的人均衣食支出金額比/r/n2018/r/n年的人均衣食支出金額高/r/n

/r/nB.2019/r/n年除醫(yī)療以外的人均消費(fèi)支出金額等于/r/n2018/r/n年的人均消費(fèi)總支出金額/r/n

/r/nC.2019/r/n年的人均文教支出比例比/r/n2018/r/n年的人均文教支出比例有提高/r/n

/r/nD.2019/r/n年人均各項(xiàng)消費(fèi)支出中，“其他”消費(fèi)支出的年增長(zhǎng)率最低/r/nACD/r/n．/r/n∵2019/r/n年居民人均消費(fèi)總支出比/r/n2018/r/n年居民人均消費(fèi)總支出提高/r/n8.5%/r/n，/r/n

/r/n∴2019/r/n年居民人均消費(fèi)總支出為：/r/n

/r/n(7000+4600+2300+1700+4400)×1.08=21600/r/n，/r/n

對(duì)于/r/nA/r/n，/r/n2019/r/n年的人均衣食支出金額為：/r/n21600×34.5%=7452/r/n元，/r/n

/r/n∴2019/r/n年的人均衣食支出金額比/r/n2018/r/n年的人均衣食支出金額高，故/r/nA/r/n正確；/r/n

對(duì)于/r/nB/r/n，/r/n2019/r/n年除醫(yī)療以外的人均消費(fèi)支出金額為：/r/n

/r/n21700×(1-8.5%)=19855.5/r/n，/r/n

/r/n2018/r/n年的人均消費(fèi)總支出金額為/r/n7000+4600+2300+1700+4400=20000/r/n元，/r/n

/r/n2019/r/n年除醫(yī)療以外的人均消費(fèi)支出金額不等于/r/n2018/r/n年的人均消費(fèi)總支出金額，故/r/nB/r/n錯(cuò)誤；/r/n

對(duì)于/r/nC/r/n，/r/n2019/r/n年的人均文教支出比例為/r/n12.0%/r/n，/r/n

/r/n2018/r/n年的人均文教支出比例為/r/n2300/r/n20000/r/n×100%=11.5%/r/n，/r/n

/r/n∴2019/r/n年的人均文教支出比例比/r/n2018/r/n年的人均文教支出比例有提高，故/r/nC/r/n正確；/r/n

對(duì)于/r/nD/r/n，/r/n2018/r/n其他支出/r/n4400/r/n元，/r/n2019/r/n年其他支出/r/n21600×21.0%=4536/r/n元，/r/n

“其他”消費(fèi)支出的年增長(zhǎng)率為/r/n4536-4400/r/n4400/r/n×100%≈3.09%/r/n，/r/n

衣食支出的年增長(zhǎng)率為：/r/n21600×34.5%-7000/r/n7000/r/n×100%≈6.46%/r/n，/r/n

住支出的年增長(zhǎng)率為：/r/n21600×24.0%-4600/r/n4600/r/n×100%≈12.70%/r/n，/r/n

文教支出的年增長(zhǎng)率為：/r/n21600×12.0%-2300/r/n2300/r/n×100%≈12.70%/r/n，/r/n

醫(yī)療支出的年增長(zhǎng)率為：/r/n21600×8.5%-1700/r/n1700/r/n×100%=8%/r/n，/r/n

/r/n∴2019/r/n年人均各項(xiàng)消費(fèi)支出中，“其他”消費(fèi)支出的年增長(zhǎng)率最低，故/r/nD/r/n正確．/r/n

故選：/r/nACD./r/n

/r/n利用條形圖和餅狀圖的性質(zhì)直接求解．/r/n

本題考查命題真假的判斷，考查條形圖、餅狀圖的性質(zhì)等基礎(chǔ)知識(shí)，考查運(yùn)算求能力、數(shù)據(jù)分析能力等數(shù)學(xué)核心素養(yǎng)，是基礎(chǔ)題．/r/nA./r/n0.8/r/n /r/nB./r/n0.625/r/n /r/nC./r/n0.5/r/n /r/nD./r/n0.1/r/nA/r/n．/r/n設(shè)發(fā)生中度霧霾為事件/r/nA/r/n，刮四級(jí)以上大風(fēng)為事件/r/nB/r/n，/r/n

所以/r/nP(A)=0.25/r/n，/r/nP(B)=0.4/r/n，/r/nP(AB)=0.2/r/n，/r/n

則在發(fā)生中度霧霾的情況下，刮四級(jí)以上大風(fēng)的概率為/r/nP(B|A)=/r/nP(AB)/r/nP(A)/r/n=/r/n0.2/r/n0.25/r/n=0.8./r/n27.(2021?寧夏中衛(wèi)三模?理T/r/n7/r/n．)已/r/n知矩形/r/nABCD/r/n的四個(gè)頂點(diǎn)的坐標(biāo)分別是/r/nA/r/n（﹣/r/n1/r/n，/r/n1/r/n），/r/nB/r/n（/r/n1/r/n，/r/n1/r/n），/r/nC/r/n（/r/n1/r/n，/r/n0/r/n），/r/nD/r/n（﹣/r/n1/r/n，/r/n0/r/n），其中/r/nA/r/n，/r/nB/r/n兩點(diǎn)在曲線/r/ny/r/n＝/r/nx/r/n2/r/n上，如圖所示．若將一枚骰子隨機(jī)放入矩形/r/nABCD/r/n中，則骰子落入陰影區(qū)域的概率是（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nC/r/n．/r/n由題意結(jié)合定積分的幾何意義可得陰影部分的面積為：/r/n，/r/n結(jié)合幾何概型計(jì)算公式可得：骰子落在陰影部分的概率為/r/n．/r/n28.(2021?寧夏中衛(wèi)三模?理T/r/n5/r/n．)/r/n2021/r/n年起，我市將試行“/r/n3+1+2/r/n”的普通高三高考新模式，即語(yǔ)文、數(shù)學(xué)、外語(yǔ)/r/n3/r/n門(mén)必選科目外，考生再?gòu)奈锢?、歷史中選/r/n1/r/n門(mén)，從化學(xué)、生物、地理、政治中選/r/n2/r/n門(mén)作為選考科目，為了幫助學(xué)生合理選科，某中學(xué)將高一每個(gè)學(xué)生的六門(mén)科目綜合成績(jī)按比例均縮放成/r/n5/r/n分制，繪制成雷達(dá)圖．甲同學(xué)的成績(jī)雷達(dá)圖如圖所示，下面敘述一定不正確的是（）/r/nA/r/n．甲的物理成績(jī)領(lǐng)先年級(jí)平均分最多/r/n /r/nB/r/n．甲有/r/n2/r/n個(gè)科目的成績(jī)低于年級(jí)平均分/r/n /r/nC/r/n．甲的成績(jī)最好的前兩個(gè)科目是化學(xué)和地理/r/n /r/nD/r/n．對(duì)甲而言，物理、化學(xué)、地理是比較理想的一種選科結(jié)果/r/nC/r/n．/r/n根據(jù)雷達(dá)圖可知甲同學(xué)物理、化學(xué)、地理成績(jī)領(lǐng)先年級(jí)平均分，其中物理、化學(xué)、地理成績(jī)領(lǐng)先年級(jí)平均分分別約為/r/n1.5/r/n分、/r/n1/r/n分，/r/n1/r/n分，所以甲同學(xué)物理成績(jī)領(lǐng)先年級(jí)平均分最多，故/r/nA/r/n項(xiàng)敘述正確，/r/nC/r/n項(xiàng)敘述不正確；/r/nB/r/n項(xiàng)：根據(jù)雷達(dá)圖可知，甲同學(xué)的歷史、政治成績(jī)低于年級(jí)平均分，故/r/nB/r/n項(xiàng)敘述正確；/r/n對(duì)甲而言，物理、化學(xué)、地理是比較理想的種選科結(jié)果，故/r/nD/r/n項(xiàng)敘述正確/r/n．/r/n29.(2021?江西南昌三模?理T/r/n7/r/n．)/r/n隨機(jī)變量/r/nX/r/n服從正態(tài)分布，有下列四個(gè)/r/n①/r/nP/r/n（/r/nX/r/n≥/r/nk/r/n）＝/r/n0.5/r/n；/r/n②/r/nP/r/n（/r/nX/r/n＜/r/nk/r/n）＝/r/n0.5/r/n；/r/n③/r/nP/r/n（/r/nX/r/n＞/r/nk/r/n+1/r/n）＜/r/nP/r/n（/r/nX/r/n＜/r/nk/r/n﹣/r/n2/r/n）；/r/n④/r/nP/r/n（/r/nk/r/n﹣/r/n1/r/n＜/r/nX/r/n＜/r/nk/r/n）＞/r/nP/r/n（/r/nk/r/n+1/r/n＜/r/nX/r/n＜/r/nk/r/n+2/r/n）．/r/n若只有一個(gè)假命題，則該假命題是（）/r/nA/r/n．/r/n①/r/n /r/nB/r/n．/r/n②/r/n /r/nC/r/n．/r/n③/r/n /r/nD/r/n．/r/n④/r/nC/r/n．/r/n因?yàn)?r/n4/r/n個(gè)命題中只有一個(gè)假命題，又/r/n①/r/nP/r/n（/r/nX/r/n≥/r/nk/r/n）＝/r/n0.5/r/n；/r/n②/r/nP/r/n（/r/nX/r/n＜/r/nk/r/n）＝/r/n0.5/r/n，/r/n由正態(tài)分布的相知可知，/r/n①②/r/n均為真命題，所以/r/nμ/r/n＝/r/nk/r/n，/r/n則/r/nP/r/n（/r/nX/r/n＞/r/nk/r/n+1/r/n）＞/r/nP/r/n（/r/nX/r/n＞/r/nk/r/n+2/r/n）＝/r/nP/r/n（/r/nX/r/n＜/r/nk/r/n﹣/r/n2/r/n），故/r/n③/r/n錯(cuò)誤；/r/n因?yàn)?r/nP/r/n（/r/nk/r/n﹣/r/n1/r/n＜/r/nX/r/n＜/r/nk/r/n）＝/r/nP/r/n（/r/nk/r/n＜/r/nX/r/n＜/r/nk/r/n+1/r/n）＞/r/nP/r/n（/r/nk/r/n+1/r/n＜/r/nX/r/n＜/r/nk/r/n+2/r/n），故/r/n④/r/n正確．/r/n30.(2021?安徽馬鞍山三模?理T/r/n3/r/n．)/r/n雷達(dá)圖也稱為網(wǎng)絡(luò)圖、蜘蛛圖，是一種能夠直觀地展示多維度的類目數(shù)據(jù)對(duì)比情況的統(tǒng)計(jì)圖．如圖是小明、小張和小陳三位同學(xué)在高一一學(xué)年六科平均成績(jī)雷達(dá)圖，則下列說(shuō)法錯(cuò)誤的是（）/r/nA/r/n．綜合六科來(lái)看，小明的成績(jī)最好，最均衡/r/n /r/nB/r/n．三人中，小陳的每門(mén)學(xué)科的平均成績(jī)都是最低的/r/n /r/nC/r/n．六門(mén)學(xué)科中，小張存在偏科情況/r/n /r/nD/r/n．小陳在英語(yǔ)學(xué)科有較強(qiáng)的學(xué)科優(yōu)勢(shì)/r/nB/r/n．/r/n對(duì)于/r/nA/r/n，小明各科成績(jī)都處于較高分?jǐn)?shù)段且圖形最均衡，由此可知小明成績(jī)最好，最均衡，故選項(xiàng)/r/nA/r/n正確；/r/n對(duì)于/r/nB/r/n，小陳的英語(yǔ)平均成績(jī)是三人中最高的，故選項(xiàng)/r/nB/r/n錯(cuò)誤；/r/n對(duì)于/r/nC/r/n，小張數(shù)學(xué)平均成績(jī)?yōu)闈M分，化學(xué)接近滿分，但物理和英語(yǔ)成績(jī)均為三人中最低，可知小張存在偏科情況，故選項(xiàng)/r/nC/r/n正確；/r/n對(duì)于/r/nD/r/n，小陳的英語(yǔ)平均成績(jī)是三人中最高且接近滿分，所以小陳在英語(yǔ)學(xué)科有較強(qiáng)的學(xué)科優(yōu)勢(shì)，故選項(xiàng)/r/nD/r/n正確．/r/n31.(2021?安徽馬鞍山三模?文T/r/n4/r/n．)/r/n第/r/n31/r/n屆世界大學(xué)生夏季運(yùn)動(dòng)會(huì)將于/r/n2021/r/n年/r/n8/r/n月在成都舉行，舉辦方將招募志愿者在賽事期間為運(yùn)動(dòng)會(huì)提供咨詢、交通引導(dǎo)、場(chǎng)館周邊秩序維護(hù)等服務(wù)．招募的志愿者需接受專業(yè)培訓(xùn)，甲、乙兩名志愿者在培訓(xùn)過(guò)程中進(jìn)行了六次測(cè)試，其測(cè)試成績(jī)（單位：分）如折線圖所示，則下列說(shuō)法正確的是（）/r/nA/r/n．甲成績(jī)的中位數(shù)比乙成績(jī)的中位數(shù)大/r/n /r/nB/r/n．甲成績(jī)的眾數(shù)比乙成績(jī)的眾數(shù)小/r/n /r/nC/r/n．甲成績(jī)的極差比乙成績(jī)的極差小/r/n /r/nD/r/n．乙的成績(jī)比甲的成績(jī)穩(wěn)定/r/nD/r/n．/r/n甲的成績(jī)分別為/r/n90/r/n，/r/n93/r/n，/r/n92/r/n，/r/n94/r/n，/r/n96/r/n，/r/n93/r/n，乙的成績(jī)分別為/r/n93/r/n，/r/n94/r/n，/r/n91/r/n，/r/n95/r/n，/r/n92/r/n，/r/n93/r/n，/r/nA/r/n：∵甲成績(jī)的中位數(shù)為/r/n＝/r/n93/r/n，乙成績(jī)的中位數(shù)為/r/n＝/r/n93/r/n，∴/r/nA/r/n錯(cuò)誤，/r/nB/r/n：∵甲成績(jī)的眾數(shù)為/r/n93/r/n，乙成績(jī)的眾數(shù)為/r/n93/r/n，∴/r/nB/r/n錯(cuò)誤，/r/nC/r/n：∵甲成績(jī)的極差為/r/n96/r/n﹣/r/n90/r/n＝/r/n6/r/n，乙成績(jī)的極差為/r/n95/r/n﹣/r/n91/r/n＝/r/n4/r/n，∴/r/nC/r/n錯(cuò)誤，/r/nD/r/n：∵甲成績(jī)的平均數(shù)為/r/n＝/r/n93/r/n，∴甲成績(jī)的方差為/r/n＝/r/n，/r/n∵乙成績(jī)的平均數(shù)為/r/n＝/r/n93/r/n，∴乙成績(jī)的方差為/r/n＝/r/n，/r/n∵/r/n＞/r/n，∴乙成績(jī)比甲成績(jī)穩(wěn)定，∴/r/nD/r/n正確．/r/n32.(2021?河北邯鄲二模?理T5/r/n．)/r/n某商場(chǎng)有三層樓，最初規(guī)劃一層為生活用品區(qū)，二層為服裝區(qū)，三層為餐飲區(qū)，招商工作結(jié)束后，共有/r/n100/r/n家商家人駐，各樓層的商鋪種類如表所示，若從所有商鋪中隨機(jī)抽取一家，該商鋪所在樓層與最初規(guī)劃不一致的概率為（）/r/n生活用品店/r/n服裝店/r/n餐飲店/r/n一層/r/n25/r/n7/r/n3/r/n二層/r/n4/r/n27/r/n4/r/n三層/r/n6/r/n1/r/n23/r/nA/r/n．/r/n0.75/r/n /r/nB/r/n．/r/n0.6/r/n /r/nC/r/n．/r/n0.4/r/n /r/nD/r/n．/r/n0.25/r/nD/r/n．/r/n100/r/n家商鋪中與最初規(guī)劃一致的有/r/n25+27+23/r/n＝/r/n75/r/n家，/r/n故不一致的有/r/n100/r/n﹣/r/n75/r/n＝/r/n25/r/n家，/r/n所以從所有商鋪中隨機(jī)抽取一家，該商鋪所在樓層與最初規(guī)劃不一致的概率為/r/n．/r/n33.(2021?河北邯鄲二模?理T/r/n3/r/n．)/r/n某校初一有/r/n500/r/n名學(xué)生，為了培養(yǎng)學(xué)生良好的閱讀習(xí)慣，學(xué)校要求他們從四大名著中選一本閱讀，其中有/r/n200/r/n人選《三國(guó)演義》，/r/n125/r/n人選《水滸傳》，/r/n125/r/n人選《西游記》，/r/n50/r/n人選《紅樓夢(mèng)》，若采用分層抽樣的方法隨機(jī)抽取/r/n40/r/n名學(xué)生分享他們的讀后感，則選《西游記》的學(xué)生抽取的人數(shù)為（）/r/nA/r/n．/r/n5/r/n /r/nB/r/n．/r/n10/r/n /r/nC/r/n．/r/n12/r/n /r/nD/r/n．/r/n15/r/nB/r/n．/r/n根據(jù)分層抽樣的定義可得選《西游記》的學(xué)生抽取的人數(shù)為/r/n×/r/n125/r/n＝/r/n10/r/n．/r/n34.(2021?江西鷹潭二模?理T/r/n9/r/n．)/r/n如圖是一個(gè)正方體紙盒的展開(kāi)圖，把/r/n1/r/n，/r/n1/r/n，/r/n2/r/n，/r/n2/r/n，/r/n3/r/n，/r/n3/r/n分別填入小正方形后，按虛線折成正方體，則所得到的正方中體相對(duì)面上的兩個(gè)數(shù)都相等的概率是（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nC/r/n．/r/n由題意，圖中有/r/n6/r/n個(gè)位置，將/r/n1/r/n，/r/n1/r/n，/r/n2/r/n，/r/n2/r/n，/r/n3/r/n，/r/n3/r/n這/r/n6/r/n個(gè)數(shù)字在/r/n6/r/n個(gè)位置全排列，共有/r/nA/r/n6/r/n6/r/n種結(jié)果，/r/n要使所得到的正方中體相對(duì)面上的兩個(gè)數(shù)都相等都相等，必須是/r/n1/r/n、/r/n1/r/n相對(duì)，/r/n2/r/n、/r/n2/r/n相對(duì)，/r/n3/r/n、/r/n3/r/n相對(duì)，/r/n正方體有/r/n6/r/n個(gè)面，寫(xiě)第一個(gè)數(shù)字時(shí)有/r/n6/r/n種選擇，/r/n剩下四個(gè)面，則第三個(gè)數(shù)字只有/r/n4/r/n種選擇，/r/n此時(shí)剩余兩個(gè)面，/r/n2/r/n個(gè)數(shù)字，有/r/n2/r/n種選擇；/r/n以此類推，可得出正方體兩個(gè)對(duì)面上兩數(shù)字和相等的組合方式有/r/n6/r/n×/r/n4/r/n×/r/n2/r/n＝/r/n48/r/n．/r/n∴所得到的正方中體相對(duì)面上的兩個(gè)數(shù)都相等的概率為：/r/nP/r/n＝/r/n＝/r/n．/r/n35.(2021?江西上饒二模?理T/r/n8/r/n．)/r/n在邊長(zhǎng)為/r/n4/r/n的正方形/r/nABCD/r/n內(nèi)部任取一點(diǎn)/r/nM/r/n，則滿足∠/r/nAMB/r/n為銳角的概率為（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nA/r/n．/r/n如果∠/r/nAEB/r/n為直角，動(dòng)點(diǎn)/r/nE/r/n位于以/r/nAB/r/n為直徑的圓上（如圖所示）．/r/n要使∠/r/nAMB/r/n為銳角，則點(diǎn)/r/nM/r/n位于正方形內(nèi)且半圓外（如圖所示的陰影部分）；/r/n因?yàn)榘雸A的面積為/r/n，正方形的面積為/r/n4/r/n×/r/n4/r/n＝/r/n16/r/n，/r/n所以滿足∠/r/nAMB/r/n為銳角的概率/r/nP/r/n＝/r/n1/r/n﹣/r/n＝/r/n1/r/n﹣/r/n．/r/n36.(2021?河北秦皇島二模?理T/r/n7/r/n．)/r/n某地病毒爆發(fā)，全省支援，需要從我市某醫(yī)院某科室的/r/n5/r/n名男醫(yī)生（含一名主任醫(yī)師）、/r/n4/r/n名女醫(yī)生（含一名主任醫(yī)師）中分別選派/r/n3/r/n名男醫(yī)生和/r/n2/r/n名女醫(yī)生，則在有一名主任醫(yī)師被選派的條件下，兩名主任醫(yī)師都被選派的概率為（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nD/r/n．/r/n需要從我市某醫(yī)院某科室的/r/n5/r/n名男醫(yī)生（含一名主任醫(yī)師）、/r/n4/r/n名女醫(yī)生（含一名主任醫(yī)師）中分別選派/r/n3/r/n名男醫(yī)生和/r/n2/r/n名女醫(yī)生，/r/n設(shè)事件/r/nA/r/n表示“有一名主任醫(yī)師被選派”，/r/nB/r/n表示“另一名主任醫(yī)師被選派”，/r/nP/r/n（/r/nA/r/n）＝/r/n+/r/n＝/r/n，/r/nP/r/n（/r/nAB/r/n）＝/r/n＝/r/n，/r/n則在有一名主任醫(yī)師被選派的條件下，兩名主任醫(yī)師都被選派的概率為：/r/nP/r/n（/r/nB/r/n|/r/nA/r/n）＝/r/n＝/r/n＝/r/n．/r/n37.(2021?北京門(mén)頭溝二模?理T10)/r/n某維修公司的四個(gè)維修點(diǎn)如圖環(huán)形分布，公司給/r/nA/r/n，/r/nB/r/n，/r/nC/r/n，/r/nD/r/n四個(gè)維修點(diǎn)某種配件各/r/n50/r/n個(gè)在使用前發(fā)現(xiàn)需要將發(fā)送給/r/nA/r/n，/r/nB/r/n，/r/nC/r/n，/r/nD/r/n四個(gè)維修點(diǎn)的配件調(diào)整為/r/n40/r/n，/r/n45/r/n，/r/n54/r/n，/r/n61/r/n，但調(diào)整只能在相鄰維修點(diǎn)間進(jìn)行，每次調(diào)動(dòng)只能調(diào)整/r/n1/r/n個(gè)配件，為完成調(diào)整，則/r/n(?)/r/nA./r/n最少需要/r/n16/r/n次調(diào)動(dòng)，有/r/n2/r/n種可行方案/r/n

/r/nB./r/n最少需要/r/n15/r/n次調(diào)動(dòng)，有/r/n1/r/n種可行方案/r/n

/r/nC./r/n最少需要/r/n16/r/n次調(diào)動(dòng)，有/r/n1/r/n種可行方案/r/n

/r/nD./r/n最少需要/r/n15/r/n次調(diào)動(dòng)，有/r/n2/r/n種可行方案/r/nA/r/n．/r/n解：根據(jù)題意，因?yàn)?r/nB/r/n、/r/nD/r/n兩處互不相鄰，所以/r/nB/r/n處至少調(diào)整/r/n5/r/n次，/r/nD/r/n處至少調(diào)整/r/n11/r/n次，故最少需要調(diào)整/r/n16/r/n次/r/n

相應(yīng)的可行方案有/r/n2/r/n種，/r/n

方案①：/r/nA/r/n調(diào)整/r/n10/r/n個(gè)給/r/nD/r/n，/r/nB/r/n調(diào)整/r/n5/r/n個(gè)給/r/nC/r/n，然后/r/nC/r/n再調(diào)整/r/n1/r/n個(gè)給/r/nD/r/n；/r/n

方案②：/r/nA/r/n調(diào)整/r/n11/r/n個(gè)給/r/nD/r/n，/r/nB/r/n調(diào)整/r/n1/r/n個(gè)給/r/nA/r/n，調(diào)整/r/n4/r/n個(gè)給/r/nC/r/n，/r/n

故選：/r/nA./r/n

/r/n根據(jù)題意，先分析兩處互不相鄰/r/nBD/r/n兩處的調(diào)整方法數(shù)目，進(jìn)而分析可得答案．/r/n

本題考查合情推理的應(yīng)用，注意認(rèn)真審題，明確題意，屬于基礎(chǔ)題．/r/n

/r/n38.(2021?江西九江二模?理T/r/n6/r/n．)恩/r/n格爾系數(shù)（/r/nEngel/r/n'/r/nsCoefficien/r/n）是食品支出總額占個(gè)人消費(fèi)支出總額的比重．居民可支配收入是居民可用于最終消費(fèi)支出和儲(chǔ)蓄的總和，即居民可用于自由支配的收入．如圖為我國(guó)/r/n2013/r/n年至/r/n2019/r/n年全國(guó)恩格爾系數(shù)和居民人均可支配收入的折線圖．/r/n給出三個(gè)結(jié)論：/r/n①/r/n恩格爾系數(shù)與居民人均可支配收入之間存在負(fù)相關(guān)關(guān)系；/r/n②/r/n一個(gè)國(guó)家的恩格爾系數(shù)越小，說(shuō)明這個(gè)國(guó)家越富裕；/r/n③/r/n一個(gè)家庭收入越少，則家庭收入中用來(lái)購(gòu)買(mǎi)食品的支出所占的比重就越?。?r/n其中正確的是（）/r/nA/r/n．/r/n①/r/n /r/nB/r/n．/r/n②/r/n /r/nC/r/n．/r/n①②/r/n /r/nD/r/n．/r/n②③/r/nC/r/n．/r/n由折線圖可知，恩格爾系數(shù)在逐年下降，/r/n居民人均可支配收入在逐年增加，故兩者之間存在負(fù)相關(guān)關(guān)系，恩格爾系數(shù)越小，/r/n居民人均可支配收入越多，經(jīng)濟(jì)越富裕，故選項(xiàng)/r/n①②/r/n正確．/r/n39/r/n.(2021/r/n?浙江麗水湖州衢州二模?/r/nT7/r/n．/r/n)/r/n設(shè)/r/n0/r/n＜/r/np/r/n＜/r/n，隨機(jī)變量/r/nξ/r/n的分布列是/r/nξ/r/n﹣/r/n1/r/n0/r/n1/r/nP/r/np/r/n﹣/r/np/r/n則當(dāng)/r/nP/r/n在（/r/n0/r/n，/r/n）內(nèi)增大時(shí)，（）/r/nA/r/n．/r/nD/r/n（/r/nξ/r/n）增大/r/n /r/nB/r/n．/r/nD/r/n（/r/nξ/r/n）減小/r/n /r/nC/r/n．/r/nD/r/n（/r/nξ/r/n）先減小后增大/r/n /r/nD/r/n．/r/nD/r/n（/r/nξ/r/n）先增大后減小/r/nD/r/n．/r/n由隨機(jī)變量/r/nξ/r/n的分布列可得，/r/nE/r/n（/r/nξ/r/n）＝﹣/r/n1/r/n?/r/np/r/n+0/r/n×/r/n+1/r/n＝/r/n，/r/n故/r/nD/r/n（/r/nξ/r/n）＝/r/n＝/r/n＝/r/n，/r/n其圖象為開(kāi)口向下的拋物線，對(duì)稱軸方程為/r/n，/r/n因?yàn)?r/n，所以/r/nD/r/n（/r/nξ/r/n）先增大后減?。?r/n40/r/n.(2021/r/n?山東濰坊二模?/r/nT12/r/n．/r/n)/r/n連/r/n接正方體每個(gè)面的中心構(gòu)成一個(gè)正八面體，甲隨機(jī)選擇此正八面體的三個(gè)頂點(diǎn)構(gòu)成三角形，乙隨機(jī)選擇此正八面體三個(gè)面的中心構(gòu)成三角形，且甲、乙的選擇互不影響，則（）/r/nA/r/n．甲選擇的三個(gè)點(diǎn)構(gòu)成正三角形的概率為/r/n /r/nB/r/n．甲選擇的三個(gè)點(diǎn)構(gòu)成等腰直角三角形的概率為/r/n /r/nC/r/n．乙選擇的三個(gè)點(diǎn)構(gòu)成正三角形的概率為/r/n /r/nD/r/n．甲選擇的三個(gè)點(diǎn)構(gòu)成的三角形與乙選擇的三個(gè)點(diǎn)構(gòu)成的三角形相似的概率為/r/nABD/r/n．/r/n甲隨機(jī)選擇的情況有/r/n種，乙隨機(jī)選擇的情況有/r/n種，/r/n對(duì)于/r/nA/r/n，甲選擇的三個(gè)點(diǎn)構(gòu)成正三角形，只有一種情況：/r/n甲從上下兩個(gè)點(diǎn)中選一個(gè)，從中間四個(gè)點(diǎn)中選相鄰兩個(gè)，共有/r/n種，/r/n故甲選擇的三個(gè)點(diǎn)構(gòu)成正三角形的概率為/r/n，故選項(xiàng)/r/nA/r/n正確；/r/n對(duì)于/r/nB/r/n，甲選擇的三個(gè)點(diǎn)構(gòu)成等腰直角三角形，有三種情況：/r/n①/r/n上下兩點(diǎn)都選，中間四個(gè)點(diǎn)中選一個(gè)，共有/r/n＝/r/n4/r/n種；/r/n②/r/n上下兩點(diǎn)鐘選一個(gè)，中間四個(gè)點(diǎn)中選相對(duì)的兩個(gè)點(diǎn)，共有/r/n種；/r/n③/r/n中間四個(gè)點(diǎn)中選三個(gè)點(diǎn)，共有/r/n種，/r/n故共有/r/n4+4+4/r/n＝/r/n12/r/n種，/r/n所以甲選擇的三個(gè)點(diǎn)構(gòu)成等腰直角三角形的概率為/r/n，故選項(xiàng)/r/nB/r/n正確；/r/n對(duì)于/r/nC/r/n，乙選擇的三個(gè)點(diǎn)構(gòu)成正三角形，只有一種情況：/r/n上面四個(gè)面的中心中選一個(gè)點(diǎn)且從下面四個(gè)面的中心選相對(duì)的兩個(gè)點(diǎn)，或下面四個(gè)面的中心中選一個(gè)點(diǎn)且從上面四個(gè)面的中心選相對(duì)的兩個(gè)點(diǎn)，共有/r/n種，/r/n所以乙選擇的三個(gè)點(diǎn)構(gòu)成正三角形的概率為/r/n，故選項(xiàng)/r/nC/r/n錯(cuò)誤；/r/n對(duì)于/r/nD/r/n，選擇的三個(gè)點(diǎn)構(gòu)成等腰直角三角形同上所求，共有/r/n8+16/r/n＝/r/n24/r/n種，概率為/r/n，/r/n甲乙相似，則甲乙均為正三角形或均為等腰直角三角形，/r/n所以甲選擇的三個(gè)點(diǎn)構(gòu)成的三角形與乙選擇的三個(gè)點(diǎn)構(gòu)成的三角形相似的概率為/r/n，故選項(xiàng)/r/nD/r/n正確．/r/n41.(2021?遼寧朝陽(yáng)二模?T11．)下/r/n列說(shuō)法正確的是（）/r/nA/r/n．將一組數(shù)據(jù)中的每個(gè)數(shù)據(jù)都乘以同一個(gè)非零常數(shù)/r/na/r/n后，方差也變?yōu)樵瓉?lái)的/r/na/r/n倍/r/nB/r/n．若四條線段的長(zhǎng)度分別是/r/n1/r/n，/r/n3/r/n，/r/n5/r/n，/r/n7/r/n，從中任取/r/n3/r/n條，則這/r/n3/r/n條線段能夠成三角形的概率為/r/n /r/nC/r/n．線性相關(guān)系數(shù)/r/nr/r/n越大，兩個(gè)變量的線性相關(guān)性越強(qiáng)；反之，線性相關(guān)性越弱/r/n /r/nD/r/n．設(shè)兩個(gè)獨(dú)立事件/r/nA/r/n和/r/nB/r/n都不發(fā)生的概率為/r/n，/r/nA/r/n發(fā)生且/r/nB/r/n不發(fā)生的概率與/r/nB/r/n發(fā)生且/r/nA/r/n不發(fā)生的概率相同，則事件/r/nA/r/n發(fā)生的概率為/r/nBD/r/n．/r/nA/r/n．將一組數(shù)據(jù)中的每個(gè)數(shù)據(jù)都乘以同一個(gè)非零常數(shù)/r/na/r/n后，方差變?yōu)樵瓉?lái)的/r/na/r/n2/r/n倍，故/r/nA/r/n錯(cuò)誤，/r/nB/r/n．從中任取/r/n3/r/n條共有/r/n4/r/n種，若三段能構(gòu)成三角形，則只有/r/n3/r/n，/r/n5/r/n，/r/n7/r/n，一種，則構(gòu)成三角形的概率是/r/n，故/r/nB/r/n正確，/r/nC/r/n．/r/n|/r/nr/r/n|/r/n→/r/n1/r/n，兩個(gè)變量的線性相關(guān)性越強(qiáng)，/r/n|/r/nr/r/n|/r/n→/r/n0/r/n，線性相關(guān)性越弱，故/r/nC/r/n錯(cuò)誤，/r/nD/r/n．由題意知/r/nP/r/n（/r/n）?/r/nP/r/n（/r/n）＝/r/n，/r/nP/r/n（/r/n）?/r/nP/r/n（/r/nB/r/n）＝/r/nP/r/n（/r/nA/r/n）?/r/nP/r/n（/r/n），/r/n設(shè)/r/nP/r/n（/r/nA/r/n）＝/r/nx/r/n，/r/nP/r/n（/r/nB/r/n）＝/r/ny/r/n，則/r/n，/r/n得/r/n得/r/nx/r/n2/r/n﹣/r/n2/r/nx/r/n+1/r/n＝/r/n，即（/r/nx/r/n﹣/r/n1/r/n）/r/n2/r/n＝/r/n，得/r/nx/r/n﹣/r/n1/r/n＝/r/n或/r/nx/r/n﹣/r/n1/r/n＝﹣/r/n，/r/n得/r/nx/r/n＝/r/n（舍）或/r/nx/r/n＝/r/n，即事件/r/nA/r/n發(fā)生的概率為/r/n，故/r/nD/r/n正確．/r/n42.(2021?遼寧朝陽(yáng)二模?T2．)某/r/n校有/r/n1000/r/n人參加某次模擬，其中數(shù)學(xué)成績(jī)近似服從正態(tài)分布/r/nN/r/n（/r/n105/r/n，σ/