有五封信随机装入五个信封,求至少有两封信与信封一值的概率

断线风筝0002022-10-04 11:39:543条回答

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鱼嘻水 共回答了21个问题 | 采纳率100%
回答:
这个问题属于“乱序”(Derangement)问题的变种.
共有5! = 120种放法.放对0封和1封的情况分别是C(5, 0)x44 = 44和C(5, 1)x9 = 45.故放对2封以及2封以上的概率是(120 - 44 - 45) / 120 = 31/120 ≈ 0.2583.
1年前
今颦眉 共回答了45个问题 | 采纳率
试求至少有两封信与信封标号一致的的概率。 还有 解出来的话能详细的说明下.谢谢 总放法为5A5=5!=120种 5个全一致情形只有1种有且仅有3个一致
1年前
梁方雨 共回答了1个问题 | 采纳率
分三类,第一类,恰有两封信与信封一致,有C(5,2)*2=20种放法;第二类,恰有三封信与信封一致,有C(5,3)*1=10种放法;第三类,五封信与信封都一致,有1种放法。
所以,所求概率为:(20+30+1)/5!=31/120。
1年前

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把五封信投入三个信箱,每个信箱至少投一封,问有几种不同的投法?
把五封信投入三个信箱,每个信箱至少投一封,问有几种不同的投法?
麻烦给我详细的分析过程.
答案是150
xiaoban83181年前11
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分完信后,各信箱的信的数量情况可能是221
或者113
221的情况,选出只有1封的信箱C3(1)
选出投到该信箱的信C5(1)
剩下的4封投2个信箱C4(2)
所以可能有3*5*6=90 种
113的情况,选出有3封的信箱C3(1)
选出投到该信箱的信C5(3)
剩下的两封投到2个箱中,C2(1)
所以可能有3*10*2=60种
90+60=150
所以有150种
(2006•陕西)某单位需以“挂号信”或“特快专递”方式向五所学校各寄一封信.这五封信的重量分别是72g,90g,215
(2006•陕西)某单位需以“挂号信”或“特快专递”方式向五所学校各寄一封信.这五封信的重量分别是72g,90g,215g,340g,400g.根据这五所学校的地址及信件的重量范围,在邮局查得相关邮费标准如下:
业务种类计费单位资费标准(元)挂号费(元/封)特制信封
(元/个)
挂号信首重100g,每重20g0.830.5
续重101~2000g,每重100g2.00
特快专递首重1000g内5.0031.0
(1)重量为90g的信若以“挂号信”方式寄出,邮寄费为多少元?若以“特快专递”方式寄出呢?
(2)这五封信分别以怎样的方式寄出最合算?请说明理由.
(3)通过解答上述问题,你有何启示?(请你用一、两句话说明)
华纸煌1年前1
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解题思路:根据表中提供的信息,对每种重量的信件的费用进行计算,选出最合理的方案.

(1)重量为90g的信以“挂号信”方式寄出,则邮寄费为5×0.8+3+0.5=7.5(元);
以“特快专递”方式寄出,邮寄费为5+3+1=9(元).

(2)∵这五封信的重量均小于1000g,
∴若以“特快专递”方式寄出,邮寄费为5+3+1=9(元).
由(1)得知,重量为90g的信以“挂号信”方式寄出,费用为7.5元小于9元;
∵72g<90g,
∴重量为72g的信以“挂号信”方式寄出小于9元;
若重量为215g的信以“挂号信”方式寄出,则
邮寄费为5×0.8+2×2+3+0.5=11.5(元)>9(元).
∵400g>340g>215g,
∴重量为400g,340g的信以“挂号信”方式寄出,费用均超过9元.
因此,将这五封信的前两封以“挂号信”方式寄出,
后三封以“特快专递”方式寄出最合算.

(3)学生言之有理即可.如:在生活中遇到花钱的问题要多计算一下,选最优方案.

点评:
本题考点: 有理数的混合运算.

考点点评: 此题信息量大,涉及很多专业术语,阅读时要弄清题意,以免算错.特别要想一想有何启示.

今有标号为1,2,3,4,5的五封信,另有同样标号的五个信封.现将五封信任意地装入五个信封,每个信封装入一封信,试求至少
今有标号为1,2,3,4,5的五封信,另有同样标号的五个信封.现将五封信任意地装入五个信封,每个信封装入一封信,试求至少有两封信配对的概率.
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解题思路:至少有两封信配对包括恰有两封信配对、恰有三封信配对、恰有五封信配对三种情况,而这三种情况对应事件为互斥事件,故分别求概率再取和即可.而每种情况对应的概率可由古典概型求解.

设恰有两封信配对为事件A,
恰有三封信配对为事件B,
恰有四封信(也即五封信配对)为事件C,
则“至少有两封信配对”事件等于A+B+C,且A、B、C两两互斥.
∵P(A)=

C25•2

A55,P(B)=

C35

A55,P(C)=[1

A55,
∴所求概率P(A)+P(B)+P(C)=
31/120].
答:至少有两封信配对的概率是[31/120].

点评:
本题考点: 互斥事件的概率加法公式;等可能事件的概率.

考点点评: 本题考查古典概型、互斥事件的概率加法、排列、组合等知识,考查分析问题、解决问题的能力.

五封信投入三个不同信箱,每一个信箱都有信,问有多少种情况
五封信投入三个不同信箱,每一个信箱都有信,问有多少种情况
如题,我的想法是分成两种情况,其一是311型的,就是C53×A33一共是六十种,其二是221型的,我认为是C52×C32×A33为180种,但这个答案显然是不对的,而就我猜测应该是221型的问题,检查一下我哪个地方出错了,
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应该是5封不同的信吧.
这是分堆问题,不论是311还是221,分成的堆中都有两堆的数是相同的,这时需要除以相同堆的全排的.
比如 311型:
(C(5,3)*C(2,1)*C(1,1)/A(2,2))*A(3,3)=60
因为有两堆相同都是1封,所以需要除以A(2,2)
同样,对于 221型:
(C(5,2)*C(3,2)*C(1,1)/A(2,2))*A(3,3)=90
22,说明有相两堆都是两封,所以需要除以 A(2,2)
这样总数是 60+90=150
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装信封一共有A(上5下5)=120种, 至少2封装对的情况为C(上2下5)*A(上3下3)=60种 所以概率为50%
五封信装入五个信封全部装错的情况有多少种?
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答案:44种
全部装法一共是120种
全部装对1种
装对4封不可能
装对3封是 C53=10种
装对2封是 C52*2=20种
装对1封是 C51*9=45种
120-45-10-20-1=44种
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其中有一个邮筒中要放两封:C(5,2),然后其他三封要有顺序P(3,3)
一共有四个邮筒,这样的情况有四种
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注:括号里的数字在前面的数字在下,后面的数字在上
今有标号为1,2,3,4,5的五封信,另有同样标号的五个信封.现将五封信任意地装入五个信封,每个信封装入一封信,试求至少
今有标号为1,2,3,4,5的五封信,另有同样标号的五个信封.现将五封信任意地装入五个信封,每个信封装入一封信,试求至少有两封信配对的概率.
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解题思路:至少有两封信配对包括恰有两封信配对、恰有三封信配对、恰有五封信配对三种情况,而这三种情况对应事件为互斥事件,故分别求概率再取和即可.而每种情况对应的概率可由古典概型求解.

设恰有两封信配对为事件A,
恰有三封信配对为事件B,
恰有四封信(也即五封信配对)为事件C,
则“至少有两封信配对”事件等于A+B+C,且A、B、C两两互斥.
∵P(A)=

C25•2

A55,P(B)=

C35

A55,P(C)=[1

A55,
∴所求概率P(A)+P(B)+P(C)=
31/120].
答:至少有两封信配对的概率是[31/120].

点评:
本题考点: 互斥事件的概率加法公式;等可能事件的概率.

考点点评: 本题考查古典概型、互斥事件的概率加法、排列、组合等知识,考查分析问题、解决问题的能力.

一道---相当难的数学题~有标号的1,2,3,4,5五封信,和同样标号的5个信封.现在把信任意装入5个信封,会出现哪些情
一道---相当难的数学题~
有标号的1,2,3,4,5五封信,和同样标号的5个信封.
现在把信任意装入5个信封,会出现哪些情形?
只有一封信配对的机会是多少?
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全排列啊,5*4*3*2*1=120 种情况.只有一封信配对:5*3*2*1=30
今有标号为1,2,3,4,5的五封信,另有同样标号的五个信封.现将五封信任意地装入五个信封,每个信封装入一封信,试求至少
今有标号为1,2,3,4,5的五封信,另有同样标号的五个信封.现将五封信任意地装入五个信封,每个信封装入一封信,试求至少有两封信配对的概率.
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解题思路:至少有两封信配对包括恰有两封信配对、恰有三封信配对、恰有五封信配对三种情况,而这三种情况对应事件为互斥事件,故分别求概率再取和即可.而每种情况对应的概率可由古典概型求解.

设恰有两封信配对为事件A,
恰有三封信配对为事件B,
恰有四封信(也即五封信配对)为事件C,
则“至少有两封信配对”事件等于A+B+C,且A、B、C两两互斥.
∵P(A)=

C25•2

A55,P(B)=

C35

A55,P(C)=[1

A55,
∴所求概率P(A)+P(B)+P(C)=
31/120].
答:至少有两封信配对的概率是[31/120].

点评:
本题考点: 互斥事件的概率加法公式;等可能事件的概率.

考点点评: 本题考查古典概型、互斥事件的概率加法、排列、组合等知识,考查分析问题、解决问题的能力.