簡單概率1問......

2007-11-24 11:43 pm
1. 擲三粒均質骰子,求下列事件的概率。

(a) 擲得至少1個「1」。

2. 擲一枚非均質硬幣得「公」的概率是0.6。若擲該硬幣五次,求下列事件的概率。

(a) 擲得5個「公」。

(b) 擲得5個「公」或5個「字」。

(c) 擲得至少1個「公」和1個「字」。

回答 (2)

2007-11-25 12:38 am
✔ 最佳答案
1 (a) 1 - P(全部都不是1字)

=1 - 5/6 x 5/6 x 5/6
=1 -125/216
=0.421 //

2(a) P( 5個公字)
= (0.6)^5
=0.078 //

2(b) P(5個公字) + P(5個字字)
=(0.6)^5 + (1-0.6)^5
=0.078 + (0.4)^5
=0.078 + 0.01
=0.088 //

2(c) P(至少1個公字和1個字)
=1-P(沒有1個是公字)- P(沒有1個是字字)
=1-P(全部都是公字)- P(全部都是字字)
=1 - (0.6)^5 - (0.4)^5
=1 - 0.088
=0.912 //

2007-11-24 16:42:03 補充:
***** 下面清楚些 *************2(c) P(至少1個公字和1個字)=1-P(沒有1個是公字)- P(沒有1個是字字)=1- P(全部都是字字)- P(全部都是公字)=1 - (0.6)^5 - (0.4)^5=1 - 0.088=0.912 //
2007-11-25 12:37 am
is it throw 3 together or throw one by one?

2007-11-24 16:59:13 補充:
2.a) P(5 heads)=0.6x0.6x0.6x0.6x0.6b)P(5 heads or 5 tails)= P(5 heads) 0.4x0.4x0.4x0.4x0.4c)P(at least 1 head and 1 tail)= 1 - [(0.6)^5 (0.4)^5]

2007-11-24 17:00:37 補充:
c)P(at least 1 head and 1 tail)= 1 - [(0.6)^5 -- (0.4)^5]

2007-11-24 17:01:28 補充:
Correct onec)P(at least 1 head and 1 tail)= 1 - [(0.6)^5 (0.4)^5]

2007-11-24 17:05:39 補充:
1 (a)P(at least one 1)=1 - (5/6)3=0.421


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