作业帮GMAT数学一题,求清晰解题思路并分析本人的解法对错,A box contains 100 balls,numbered from 1 to 100.If three balls are selected at random and with replacement from the box,what is the probability that the sum of the three numbers on
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/23 22:03:53
GMAT数学一题,求清晰解题思路并分析本人的解法对错,A box contains 100 balls,numbered from 1 to 100.If three balls are selected at random and with replacement from the box,what is the probability that the sum of the three numbers on
GMAT数学一题,求清晰解题思路并分析本人的解法对错,
A box contains 100 balls,numbered from 1 to 100.If three balls are selected at random and with replacement from the box,what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
(A) 1/4
(B) 3/8
(C) 1/2
(D) 5/8
(E) 3/4
这题网上能搜到解答,但是我觉得不太正确.就算结果只有奇偶两种也有可能结果为奇的三数之和权重大于结果为偶的三数之和的权重.
我的解法是两种情况可以,1.三奇数.2.两偶一奇.
故：[C(50,3)+C(50,1)C(50,2)]/C(100,3)
麻烦帮我分析下,
GMAT数学一题,求清晰解题思路并分析本人的解法对错,A box contains 100 balls,numbered from 1 to 100.If three balls are selected at random and with replacement from the box,what is the probability that the sum of the three numbers on
这道题的解题关键在于你对“with replacement"的理解.
with replacement表示each member may be chosen more than once,就是放回抽样,从口袋里取球,第一次取一个球,观察号码后放回袋中,搅匀后再取一个,抽第一个、抽第二个、抽第三个是相互独立的.
without replacement表示each member cannot be chosen more than once,就是不放回抽样,取第一个球不放回,第二个从剩下的球中再取.
显然,这道题中你错在把with replacement当成without replacement来理解了.
解题：
解法一：考虑每个球的奇偶性
3奇：1/2*1/2*1/2*=1/8
2偶1奇：把取三个小球看成,在3个空位中取1个空位放奇数球、剩下两个放偶数球,则概率就是C（3,1）*1/2*1/2*1/2=3/8
1/8+3/8=1/2
解法二：本题中,奇偶个数相同,随机抽取且是放回抽样,所以三个球的号码和为奇数,与三个球的号码和为偶数是等可能事件,所以是1/2