2021年2019数学新设计北师大选修2-3精练第二章概率2.1.pdf
-1-第二章DIERZHANG概率1离散型随机变量及其分布列A 组1.袋中有大小、形状、质地相同的5 个球,分别标有1,2,3,4,5 五个号码,现在在有放回抽取的条件下依次取出两个球,设两个球号码之和为随机变量X,则 X 所有可能取值的个数是()A.5 B.9 C.10 D.25 解析:X 的所有可能取值为2,3,4,5,6,7,8,9,10,共 9 个.答案:B 2.设某项试验的成功率是失败率的2 倍,用随机变量X去描述 1 次试验的成功次数,则 P(X=0)等于()A.0 B.C.D.解析:设 X 的分布列为X01Pp2p即“X=0”表示试验失败,“X=1”表示试验成功,设失败率为p,则成功率为2p.由 p+2p=1,得 p=,故选 C.答案:C 3.设随机变量Y的分布列为Y-123Pm则“Y”的概率为()A.B.C.D.解析:依题意知,+m+=1,则 m=.故 P=P(Y=2)+P(Y=3)=.答案:C|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 1 页，共 5 页-2-4.抛掷两枚骰子一次,为第一枚骰子掷出的点数与第二枚骰子掷出的点数之差,则 的所有可能的取值为()A.0 5,NB.-5 0,ZC.1 6,ND.-5 5,Z解析:的所有可能取值为-5,-4,-3,-2,-1,0,1,2,3,4,5,即-5 5,Z.答案:D 5.设 X 是一个离散型随机变量,其分布列如下:X-101P1-2aa2则 a 等于()A.1 B.1C.1+D.1-解析:由分布列性质可得解得 a=1-,故选 D.答案:D 6.在 8 件产品中,有 3 件次品,5 件正品,从中任取 3 件,记次品的件数为,则 2 表示的试验结果是.解析:应分 =0 和 =1 两类.=0 表示取到 3 件正品;=1 表示取到1 件次品、2 件正品.故 2 表示的试验结果为取到 1 件次品、2 件正品或取到3 件正品.答案:取到 1 件次品、2 件正品或取到3 件正品7.设随机变量X 的分布列为P(X=k)=,k=1,2,3,其中 c 为常数,则 P(0X 2)=.解析:由题意得=1,解得 c=.所以 P(0X 6)=.解析:由随机变量分布列的性质知概率之和为1,故P(X=2)+P(X=4)+P(X=6)+P(X=8)+P(X=10)=a+2a+3a+4a+5a=15a=1,a=,P(X=2k)=k(k=1,2,3,4,5),P(X6)=P(X=8)+P(X=10)=.答案:4.若随机变量的分布列为-2-10123P0.10.20.20.30.10.1则当 P(x)=0.8 时,实数 x 的取值范围是()A.x2 B.1 x2 C.1x2 D.1x 2 解析:由随机变量 的分布列知:P(-1)=0.1,P(0)=0.3,P(1)=0.5,P(2)=0.8,则当 P(x)=0.8 时,实数 x 的取值范围是1x2.答案:C 5.若 P(x2)=1-,P(x1)=1-,其中 x1x2,则 P(x1 x2)等于.解析:由分布列性质可有:P(x1 x2)=P(x2)+P(x1)-1=(1-)+(1-)-1=1-(+).答案:1-(+)6.导学号43944028设离散型随机变量X 的分布列为|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 4 页，共 5 页文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5-5-X01234P0.20.10.10.3m求:(1)2X+1 的分布列;(2)|X-1|的分布列.解由分布列的性质知0.2+0.1+0.1+0.3+m=1,解得 m=0.3.首先列表为X012342X+113579|X-1|10123从而由上表得两个分布列为:(1)2X+1 的分布列2X+113579P0.20.10.10.30.3(2)|X-1|的分布列|X-1|0123P0.10.30.30.3|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 5 页，共 5 页文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 ZV7X8B6F10P5文档编码:CX2G1V8L2R2 HO4K6D5M5A1 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