{"id":2128,"date":"2023-04-02T09:54:33","date_gmt":"2023-04-02T01:54:33","guid":{"rendered":"https:\/\/www.appblog.cn\/?p=2128"},"modified":"2023-04-05T20:41:49","modified_gmt":"2023-04-05T12:41:49","slug":"high-school-mathematics-foundation-binomial-distribution-and-binomial-theorem","status":"publish","type":"post","link":"https:\/\/www.appblog.cn\/index.php\/2023\/04\/02\/high-school-mathematics-foundation-binomial-distribution-and-binomial-theorem\/","title":{"rendered":"\u9ad8\u4e2d\u6570\u5b66\u57fa\u7840\uff1a\u4e8c\u9879\u5206\u5e03\u4e0e\u4e8c\u9879\u5f0f\u5b9a\u7406"},"content":{"rendered":"

\u72ec\u7acb\u91cd\u590d\u8bd5\u9a8c<\/h2>\n

\u72ec\u7acb\u91cd\u590d\u8bd5\u9a8c\u7684\u57fa\u672c\u7279\u5f81\uff1a
\n1\u3001\u6bcf\u6b21\u8bd5\u9a8c\u662f\u5728\u540c\u6837\u6761\u4ef6\u4e0b\u8fdb\u884c
\n2\u3001\u6bcf\u6b21\u8bd5\u9a8c\u90fd\u662f\u53ea\u6709\u4e24\u79cd\u7ed3\u679c\uff1a\u53d1\u751f\u4e0e\u4e0d\u53d1\u751f
\n3\u3001\u5404\u6b21\u8bd5\u9a8c\u4e2d\u7684\u4e8b\u4ef6\u662f\u76f8\u4e92\u72ec\u7acb\u7684
\n4\u3001\u6bcf\u6b21\u8bd5\u9a8c\uff0c\u67d0\u4e8b\u4ef6\u53d1\u751f\u7684\u6982\u7387\u662f\u76f8\u540c\u7684<\/p>\n

<\/p>\n

\u4e8c\u9879\u5206\u5e03<\/h2>\n

\u5728$n$\u6b21\u72ec\u7acb\u91cd\u590d\u8bd5\u9a8c\u4e2d\uff0c\u8bbe\u4e8b\u4ef6$A$\u53d1\u751f\u7684\u6b21\u6570\u662f$X$\uff0c\u4e14\u5728\u6bcf\u6b21\u8bd5\u9a8c\u4e2d\u4e8b\u4ef6$A$\u53d1\u751f\u7684\u6982\u7387\u662f$p$\uff0c\u90a3\u4e48\u4e8b\u4ef6$A$\u6070\u597d\u53d1\u751f$k$\u6b21\u7684\u6982\u7387\u4e3a<\/p>\n

$P(X=k)=C_n^kp^k(1-p)^{n-k}, \\, k=0,1,2,..,n$<\/p>\n\n\n\n\n\n
$X$<\/th>\n0<\/th>\n1<\/th>\n$\u00b7\u00b7\u00b7$<\/th>\n$k$<\/th>\n$\u00b7\u00b7\u00b7$<\/th>\n$n$<\/th>\n<\/tr>\n<\/thead>\n
$P$<\/td>\n$C_n^0p^0(1-p)^n$<\/td>\n$C_n^1p^1(1-p)^{n-1}$<\/td>\n$\u00b7\u00b7\u00b7$<\/td>\n$C_n^kp^k(1-p)^{n-k}$<\/td>\n\u00b7\u00b7\u00b7<\/td>\n$C_n^np^n(1-p)^0$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\u79f0\u968f\u673a\u53d8\u91cf$X$\u670d\u4ece\u4e8c\u9879\u5206\u5e03\uff0c\u8bb0\u4f5c<\/p>\n

$X \\sim B(n,p)$\uff0c\u5176\u4e2d$p$\u4e3a\u6210\u529f\u6982\u7387\u3002<\/p>\n

\u4e8c\u9879\u5f0f\u5b9a\u7406<\/h2>\n

$$
\n\\sum_{k=0}^nC_n^kp^k(1-p)^{n-k}=(p+q)^n=1, \\, (q=1-p)
\n$$<\/p>\n

\u4e8c\u9879\u5206\u5e03\u793a\u4f8b<\/h2>\n

\u8bbe\u4e00\u5c04\u624b\u5e73\u5747\u6bcf\u5c04\u51fb10\u6b21\u4e2d\u97764\u6b21\uff0c\u6c42\u5728\u4e94\u6b21\u5c04\u51fb\u4e2d\uff0c\u2460 \u51fb\u4e2d\u4e00\u6b21\uff0c\u2461 \u7b2c\u4e8c\u6b21\u51fb\u4e2d\uff0c\u2462 \u51fb\u4e2d\u4e24\u6b21\uff0c\u2463 \u7b2c\u4e8c\u3001\u4e09\u4e24\u6b21\u51fb\u4e2d\uff0c\u2464 \u81f3\u5c11\u51fb\u4e2d\u4e00\u6b21\u7684\u6982\u7387\u3002<\/p>\n

\u7531\u9898\u8bbe\uff0c\u6b64\u5c04\u624b\u5c04\u51fb1\u6b21\uff0c\u4e2d\u9776\u7684\u6982\u7387\u4e3a$0.4$<\/p>\n

\u2460 $n=5, k=1$\uff0c\u5e94\u7528\u516c\u5f0f\u5f97<\/p>\n

$P(X=1)=C_5^1p(1-p)^4=0.2592$<\/p>\n

\u2461 \u4e8b\u4ef6\u201c\u7b2c\u4e8c\u6b21\u51fb\u4e2d\u201d\u8868\u793a\u7b2c\u4e00\u3001\u4e09\u3001\u56db\u3001\u4e94\u6b21\u51fb\u4e2d\u6216\u51fb\u4e0d\u4e2d\u90fd\u53ef\uff0c\u5b83\u4e0d\u540c\u4e8e\u201c\u51fb\u4e2d\u4e00\u6b21\u201d\uff0c\u4e5f\u4e0d\u540c\u4e8e\u201c\u7b2c\u4e8c\u6b21\u51fb\u4e2d\uff0c\u5176\u4ed6\u5404\u6b21\u90fd\u4e0d\u4e2d\u201d\uff0c\u4e0d\u80fd\u7528\u516c\u5f0f\u3002\u5b83\u7684\u6982\u7387\u5c31\u662f0.4\u3002<\/p>\n

\u2462 $n=5, k=2$\uff0c\u5e94\u7528\u516c\u5f0f\u5f97<\/p>\n

$P(X=2)=C_5^2p(1-p)^3=0.3456$<\/p>\n

\u2463 \u201c\u7b2c\u4e8c\u3001\u4e09\u4e24\u6b21\u51fb\u4e2d\u201d\u8868\u793a\u7b2c\u4e00\u6b21\u3001\u7b2c\u56db\u6b21\u53ca\u7b2c\u4e94\u6b21\u53ef\u4e2d\u53ef\u4e0d\u4e2d\uff0c\u6240\u4ee5\u6982\u7387\u4e3a$0.4\u00d70.4=0.16$\u3002<\/p>\n

\u2464 \u8bbe\u201c\u81f3\u5c11\u51fb\u4e2d\u4e00\u6b21\u201d\u4e3a\u4e8b\u4ef6$B$\uff0c\u5219$B$\u5305\u62ec\u201c\u51fb\u4e2d\u4e00\u6b21\u201d\uff0c\u201c\u51fb\u4e2d\u4e24\u6b21\u201d\uff0c\u201c\u51fb\u4e2d\u4e09\u6b21\u201d\uff0c\u201c\u51fb\u4e2d\u56db\u6b21\u201d\uff0c\u201c\u51fb\u4e2d\u4e94\u6b21\u201d\uff0c\u6240\u4ee5\u6982\u7387\u4e3a<\/p>\n

$P(B)=P_5(1)+P_5(2)+P_5(3)+P_5(4)+P_5(5)=1-P_5(0)=0.92224$<\/p>\n

\u4e8c\u9879\u5206\u5e03\u4e0e\u4e24\u70b9\u5206\u5e03\u7684\u5173\u7cfb<\/h2>\n

\u4f8b\uff1a\u7bee\u7403\u6bd4\u8d5b\u4e2d\u6bcf\u6b21\u7f5a\u7403\u547d\u4e2d\u5f971\u5206\uff0c\u4e0d\u4e2d\u5f970\u5206\uff0c\u67d0\u7bee\u7403\u8fd0\u52a8\u5458\u7684\u7f5a\u7403\u547d\u4e2d\u7387\u4e3a0.7\uff0c\u5728\u67d0\u6b21\u6bd4\u8d5b\u4e2d\u4ed6\u4e00\u5171\u7f5a\u740320\u6b21\uff0c\u5219<\/p>\n

\uff081\uff09\u4e00\u6b21\u7f5a\u7403\u7684\u5f97\u5206\u670d\u4ece\u4e24\u70b9\u5206\u5e03
\n\uff082\uff09\u547d\u4e2d\u6b21\u6570$X$\u670d\u4ece\u4e8c\u9879\u5206\u5e03\uff0c\u5373$X \\sim B(20,0.7)$<\/p>\n

\u5373\u4e24\u70b9\u5206\u5e03\u662f\u7279\u6b8a\u7684\u4e8c\u9879\u5206\u5e03$X \\sim B(1,p)$<\/p>\n

\u4e8c\u9879\u5206\u5e03\u4e0e\u8d85\u51e0\u4f55\u5206\u5e03\u7684\u5173\u7cfb<\/h2>\n

\u4f8b\uff1a\u4e00\u4e2a\u888b\u4e2d\u653e\u6709$M$\u4e2a\u7ea2\u7403\uff0c$(N-M)$\u4e2a\u767d\u7403\uff0c\u4f9d\u6b21\u4ece\u888b\u4e2d\u53d6$n$\u4e2a\u7403\uff0c\u8bb0\u4e0b\u7ea2\u7403\u7684\u4e2a\u6570$X$<\/p>\n

\uff081\uff09\u5982\u679c\u662f\u6709\u653e\u56de\u5730\u53d6\uff0c\u5219$X$\u670d\u4ece\u4e8c\u9879\u5206\u5e03<\/p>\n

$$
\nX \\sim B(n,\\frac{M}{N})
\n$$<\/p>\n

\uff082\uff09\u5982\u679c\u662f\u4e0d\u653e\u56de\u5730\u53d6\uff0c\u5219$X$\u670d\u4ece\u8d85\u51e0\u4f55\u5206\u5e03<\/strong><\/p>\n

$$
\nP(X=k)=\\frac{CM^kC<\/em>{N-M}^{n-k}}{C_N^n} \\, (k=0,1,2,\u00b7\u00b7\u00b7,m)
\n$$<\/p>\n

\u5176\u4e2d$m=min(M,n)$<\/p>\n

\u4f2f\u52aa\u5229\u5206\u5e03<\/h2>\n

\u4f2f\u52aa\u5229\u5206\u5e03(Bernoulli distribution)\u53c8\u540d\u4e24\u70b9\u5206\u5e03\u6216$0-1$\u5206\u5e03\uff0c\u4ecb\u7ecd\u4f2f\u52aa\u5229\u5206\u5e03\u524d\u9996\u5148\u9700\u8981\u5f15\u5165\u4f2f\u52aa\u5229\u8bd5\u9a8c(Bernoulli trial)\u3002<\/p>\n

\u4f2f\u52aa\u5229\u8bd5\u9a8c\u662f\u53ea\u6709\u4e24\u79cd\u53ef\u80fd\u7ed3\u679c\u7684\u5355\u6b21\u968f\u673a\u8bd5\u9a8c\uff0c\u5373\u5bf9\u4e8e\u4e00\u4e2a\u968f\u673a\u53d8\u91cf$X$\u800c\u8a00\uff1a\u4f2f\u52aa\u5229\u8bd5\u9a8c\u90fd\u53ef\u4ee5\u8868\u8fbe\u4e3a\u201c\u662f\u4e0e\u5426\u201d\u7684\u95ee\u9898\u3002<\/p>\n

\u4f8b\u5982\uff0c\u629b\u4e00\u6b21\u786c\u5e01\u662f\u6b63\u9762\u5411\u4e0a\u5417\uff1f\u521a\u51fa\u751f\u7684\u5c0f\u5b69\u662f\u4e2a\u5973\u5b69\u5417\u7b49\u7b49\u3002\u5982\u679c\u8bd5\u9a8c$E$\u662f\u4e00\u4e2a\u4f2f\u52aa\u5229\u8bd5\u9a8c\uff0c\u5c06$E$\u72ec\u7acb\u91cd\u590d\u5730\u8fdb\u884c$n$\u6b21\uff0c\u5219\u79f0\u8fd9\u4e00\u4e32\u91cd\u590d\u7684\u72ec\u7acb\u8bd5\u9a8c\u4e3a$n$\u91cd\u4f2f\u52aa\u5229\u8bd5\u9a8c\u3002\u8fdb\u884c\u4e00\u6b21\u4f2f\u52aa\u5229\u8bd5\u9a8c\uff0c\u6210\u529f$(X=1)$\u6982\u7387\u4e3a$p(0\u2264p\u22641)$\uff0c\u5931\u8d25$(X=0)$\u6982\u7387\u4e3a$1-p$\uff0c\u5219\u79f0\u968f\u673a\u53d8\u91cf$X$\u670d\u4ece\u4f2f\u52aa\u5229\u5206\u5e03\u3002\u4f2f\u52aa\u5229\u5206\u5e03\u5f0f\u79bb\u6563\u578b\u6982\u7387\u5206\u5e03\u3002<\/p>\n

\u4e8c\u9879\u5b9a\u7406<\/h2>\n

\uff081\uff09\u4e8c\u9879\u5c55\u5f00\u516c\u5f0f<\/p>\n

$$
\n(a+b)^n=C_n^0a^n+C_n^1a^{n-1}b+C_n^2a^{n-2}b^2+\u00b7\u00b7\u00b7+C_n^ra^{n-r}b^r+\u00b7\u00b7\u00b7+C_n^nb^n \\, (n \\in N)
\n$$<\/p>\n

\uff082\uff09\u9879\u7684\u7cfb\u6570\u4e0e\u4e8c\u9879\u5f0f\u7cfb\u6570<\/p>\n

\u9879\u7684\u7cfb\u6570\u4e0e\u4e8c\u9879\u5f0f\u7cfb\u6570\u662f\u4e0d\u540c\u7684\u4e24\u4e2a\u6982\u5ff5\uff0c\u4f46\u5f53\u4e8c\u9879\u5f0f\u7684\u4e24\u4e2a\u9879\u7684\u7cfb\u6570\u90fd\u4e3a1\u65f6\uff0c\u7cfb\u6570\u5c31\u662f\u4e8c\u9879\u5f0f\u7cfb\u6570\u3002\u5982\u5728$(ax+b)^n$\u7684\u5c55\u5f00\u5f0f\u4e2d\uff0c\u7b2c$(r+1)$\u9879\u7684\u4e8c\u9879\u5f0f\u7cfb\u6570\u4e3a$C_n^r$\uff0c\u7b2c$(r+1)$\u9879\u7684\u7cfb\u6570\u4e3a$C_n^ra^{n-r}b^r$\uff1b\u800c$(x+\\frac{1}{x})^n$\u7684\u5c55\u5f00\u5f0f\u4e2d\u7684\u7cfb\u6570\u7b49\u4e8e\u4e8c\u9879\u5f0f\u7cfb\u6570\u3002\u4e8c\u9879\u5f0f\u7cfb\u6570\u4e00\u5b9a\u4e3a\u6b63\uff0c\u800c\u9879\u7684\u7cfb\u6570\u4e0d\u4e00\u5b9a\u4e3a\u6b63<\/p>\n

\uff083\uff09$(1+x)^n$\u7684\u5c55\u5f00\u5f0f<\/p>\n

$$
\n(1+x)^n=C_n^0x^n+C_n^1x^{n-1}+C_n^2x^{n-2}+\u00b7\u00b7\u00b7+C_n^nx^0
\n$$<\/p>\n

\u82e5\u4ee4$x=1$\uff0c\u5219\u6709<\/p>\n

$$
\n(1+1)^n=C_n^0+C_n^1+C_n^2+\u00b7\u00b7\u00b7+C_n^n=2^n
\n$$<\/p>\n

\u82e5\u4ee4$x=-1$\uff0c\u5219\u6709\u4e8c\u9879\u5f0f\u5947\u6570\u9879\u7cfb\u6570\u7684\u548c\u7b49\u4e8e\u4e8c\u9879\u5f0f\u5076\u6570\u9879\u7cfb\u6570\u7684\u548c\u3002\u5373<\/p>\n

$$
\nC_n^0+C_n^2+\u00b7\u00b7\u00b7=C_n^1+C_n^3+\u00b7\u00b7\u00b7=2^{n-1}
\n$$<\/p>\n

\uff084\uff09\u4e8c\u9879\u5f0f\u7cfb\u6570\u7684\u6027\u8d28<\/p>\n

\u2460 \u5bf9\u79f0\u6027\uff1a\u4e0e\u9996\u5c3e\u4e24\u7aef\u201c\u7b49\u8ddd\u79bb\u201d\u7684\u4e24\u4e2a\u4e8c\u9879\u5f0f\u7cfb\u6570\u76f8\u7b49\uff0c\u5373$C_n^m=C_n^{n-m}$<\/p>\n

\u2461 \u589e\u51cf\u6027\u4e0e\u6700\u5927\u503c\uff1a\u5f53$r\u2264(n+1)\/2$\u65f6\uff0c\u4e8c\u9879\u5f0f\u7cfb\u6570$C_n^r$\u7684\u503c\u9010\u6e10\u589e\u5927\uff1b\u5f53$r\u2265(n+1)\/2$\u65f6\uff0c$C_n^r$\u7684\u503c\u9010\u6e10\u51cf\u5c0f\uff0c\u4e14\u5728\u4e2d\u95f4\u53d6\u7684\u6700\u5927\u503c\u3002\u5f53$n$\u4e3a\u5076\u6570\u65f6\uff0c\u4e2d\u95f4\u4e00\u9879\uff08\u7b2c$n\/2+1$\u9879\uff09\u7684\u4e8c\u9879\u5f0f\u7cfb\u6570$C_n^{n\/2}$\u53d6\u5f97\u6700\u5927\u503c\u3002\u5f53$n$\u4e3a\u5947\u6570\u65f6\uff0c\u4e2d\u95f4\u4e24\u9879\uff08\u7b2c$(n+1)\/2$\u548c$(n+1)\/2+1$\u9879\uff09\u7684\u4e8c\u9879\u5f0f\u7cfb\u6570$C_n^{(n-1)\/2}=C_n^{(n+1)\/2}$\u76f8\u7b49\u5e76\u540c\u65f6\u53d6\u5f97\u6700\u5927\u503c\u3002<\/p>\n

\uff085\uff09\u7cfb\u6570\u6700\u5927\u9879\u7684\u6c42\u6cd5<\/p>\n

\u8bbe\u7b2c$r$\u9879\u7684\u7cfb\u6570$A_r$\u6700\u5927\uff0c\u7531\u4e0d\u7b49\u5f0f\u7ec4<\/p>\n

$$
\n\\left \\{
\n\\begin{array} {c}
\nAr\u2265A<\/em>{r-1} \\
\nAr\u2265A<\/em>{r+1}
\n\\end{array}
\n\\right.
\n$$<\/p>\n

\u53ef\u786e\u5b9a$r$<\/p>\n

\uff085\uff09\u8d4b\u503c\u6cd5<\/p>\n

\u8bbe$f(x)=(ax+b)^n=a_0+a_1x+a_2x^2+\u00b7\u00b7\u00b7+a_nx^n$\uff0c\u6709<\/p>\n

\u2460 $a_0=f(0)$
\n\u2461 $a_0+a_1+a_2+\u00b7\u00b7\u00b7+a_n=f(1)$
\n\u2462 $a_0-a_1+a_2-a_3+\u00b7\u00b7\u00b7+(-1)^na_n=f(-1)$
\n\u2463 $a_0+a_2+a_4+a_6+\u00b7\u00b7\u00b7=(f(1)+f(-1))\/2$
\n\u2464 $a_1+a_3+a_5+a_7+\u00b7\u00b7\u00b7=(f(1)-f(-1))\/2$<\/p>\n","protected":false},"excerpt":{"rendered":"

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