201699544 2017-2-16 06:04 PM

問:

在抽30次中, 不可抽中4個目標或以上及在30個抽10裡面包括目標的概率是多少??

唸左好疚, 都唸唔到答案

請提供計算過程, 謝謝

ppresent 2017-2-16 11:33 PM

If the probability doesn't change each time sampling a target (keeps constant to be 0.1), then obviously the # of targets obtained in the 30 items (call it X1) follows a binomial distribution of B(30,0.1), just use the its pmf to find the answer to the first probability.

(hint: pmf f(x) of B(n,p) is (nCx)(p^x)(1-p)^(n-x)), required probability =

f(0)+f(1)+f(2)+f(3))

The second probability is a bit harder.

Firstly, the number of targets in the 10 items selected from the 30 items (call it X2) follows a hypergeometric distribution with replacement, its pmf f(k) is

(n1Ck)(N-n1)C(n-k)/ NCn,

where N is the total number of objects that can be chosen, ie N=30,

n is the number of objects drawn, ie n=10,

n1 is the number of targets in N items,

however here n1 is just X1, which follows a B(30,0.1) as explained above. The pmf of X2 can be found using the law of total probability, P(X2)=ΣP(X2|X1)P(X1) , summing over X1 = 0,1,...30

and the distribution of X2 can eventually be found to be also binomial (proof omitted here because too long), which is B(10,0.1) in fact.

Therefore the second required probability is just P(X2>=1)

=1-P(X2=0)

=1-0.9^10

[[i] 本帖最後由 ppresent 於 2017-2-28 09:48 PM 編輯 [/i]]

(hint: pmf f(x) of B(n,p) is (nCx)(p^x)(1-p)^(n-x)), required probability =

f(0)+f(1)+f(2)+f(3))

The second probability is a bit harder.

Firstly, the number of targets in the 10 items selected from the 30 items (call it X2) follows a hypergeometric distribution with replacement, its pmf f(k) is

(n1Ck)(N-n1)C(n-k)/ NCn,

where N is the total number of objects that can be chosen, ie N=30,

n is the number of objects drawn, ie n=10,

n1 is the number of targets in N items,

however here n1 is just X1, which follows a B(30,0.1) as explained above. The pmf of X2 can be found using the law of total probability, P(X2)=ΣP(X2|X1)P(X1) , summing over X1 = 0,1,...30

and the distribution of X2 can eventually be found to be also binomial (proof omitted here because too long), which is B(10,0.1) in fact.

Therefore the second required probability is just P(X2>=1)

=1-P(X2=0)

=1-0.9^10

[[i] 本帖最後由 ppresent 於 2017-2-28 09:48 PM 編輯 [/i]]

ppresent 2017-2-17 08:06 AM

[quote]原帖由 [i]ppresent[/i] 於 2017-2-16 11:33 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=456520184&ptid=26451081][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

...using total law of probability,... [/quote]

using the law of total probability

...using total law of probability,... [/quote]

using the law of total probability

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