Tanya prepared 4 different letters to be sent to 4 different addresses.For each letter,she prepared an envelope with its correct address.If the 4 letters are to be put into the 4 envelopes at random.What is the probability that only 1 letter will be put into the envelope with its correct address?
A 1/24
B 1/8
C 1/4
D 1/3
E 3/8
ANSWER: D 1/3 ,WHY?
4×2×1/P(4,4)=8/24=1/3
解释起来是这样的:(我个人的观点)
按步骤走(所以要相乘),先计算按题目要求方式选择有多少种可能:
1,四个信封当中有一个是正确的,可能性为4种
2,剩余三个信封中,第一个要选错,应该为2(因为剩余三封信,只有两封可以被错误的塞到这个信封)
以此类推,余下两个信封分别是1,1
总共有P(4,4)种可能性就是4×3×2×1
最后结果就是8/24=1/3
设信封a,b,c,d,信1,2,3,4,相互对应算正确(如:1对a).
总共有P4种方法.
只有1封正确的方法:1st.先选哪一封是正确的,有C(4,1)种
2st.设a-1正确,那么剩下的都不能正确.问题在于此时b不能对2了,它只能对3或4,即C(2,1)种方法.
3rd.剩下的因不能放对位置,所以没得选择.
故概率为:C(4,1)×C(2,1)/P4=1/3
强烈感谢Elynn&railkey!真是好人啊!
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