两题P的问题

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(1) A family has three children.  Assume that each child has an equal probability of being a
boy or a girl.  Also assume that the genders of the individual children are independent.  
Find the probability  P[X and Y], where X is the event that the oldest child is a girl, and
Y is the event that, among the three children, there are exactly two girls.  

(2) A population consists of people of three different types:  A, B, and C.  There are twice as
many Type A people as Type B people, and twice as many Type B people as Type C
people.  The probability of having a claim during a given year is 0.50 for Type A people,
0.30 for Type B people, and 0.10 for Type C people.  Each person will either have a
claim, or have no claim.  An insurer picks a person at random from the population, and
observes that, after one year, this person had no claim.  Find the probability that this is a
Type C person.

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关键词：Probability Independent Population Individual Different individual different children exactly family

1)
P(X)= P(oldest child is a girl)= 1/2
P(exactly two girls among three children):
GGB
GBG
BGG
[(1/2)^3]*3

but we do not need the combination, BGG

Therefore, P(X and Y)= [(1/2)^3]*2 = 1/4

2)
1B = 2A
1C = 2B
Therefore, 1C = 2B = 4A

Probability of no claim:
A: 0.50
B: 0.70
C: 0.90

Bayes' theorem:
(1/7 *0.90)/ [(1/7 *0.90)+ (2/7 *0.70)+ (4/7 *0.50)]

Compute the above and then you will get the answer.

昨天刚过的P，呵呵，趁热回答一下：
（1）符合条件的情况只有两种
GGB GBG(假设第一个字母表示最大的孩子)
而三个小孩总共有2^3=8种情况，
所以P=2/8=1/4

(2)这是一道条件概率和全概率公式的题
由题有
P(A)=2P(B)
P(B)=2P(C)
P(A)+P(B)+ P(C)=1
所以P(A)=4/7,P(B)=2/7,P(C)=1/7

接下来就和楼上一样了

PS：这两道题目应该不难吧。。。楼主要好好加油了，不知道你什么时候考，是3月吗？