1.生產一種零件,甲車間的合格率是96%,乙車間的合格率是97%,從它們生產的零件中各抽取1件,都抽取合格品的概率是多少?
2.在某段時間內,甲地下雨的概率是0.2,乙地下雨的概率是0.3,假定在這段時間內兩地是否下雨相互之間沒有影響,計算在這段時1內:
(1)甲,乙兩地都下雨的概率;
(2)甲,乙兩地都不下雨的概率;
(3)其中至少一個地方下雨的概率
3.某射手射擊1次,擊中目標的概率是0.9,他連續射擊4次,且各次射擊是否擊中相互之間沒有影響,那麼他第2次未擊﹑其他3次都擊中的概率是多少?
中5數學 相互獨立事件同時發生的概率
2007-07-21 8:03 pm
回答 (2)
2007-07-21 11:33 pm
✔ 最佳答案
1. P(都抽取合格品)= P(零件A合格) X P(零件B合格)
= 96 / 100 X 97 / 100
= 582 / 625
2. (1). P(甲、乙兩地都下雨)
= P(甲下雨) X P(乙下雨)
= 0.2 X 0.3
= 0.06
= 3 / 50
(2). P(甲、乙兩地都不下雨)
= [1 - P(甲下雨)] X [1 - P(乙下雨)]
= (1 - 0.2) X (1 - 0.3)
= 0.8 X 0.7
= 0.56
= 14 / 25
(3). P(其中至少一個地方下雨)
= 1 - P(甲、乙兩地都不下雨)
= 1 - 14 / 25
= 11 / 25
3. P(未撃中目標)
= 1 - P(擊中目標)
= 1 - 0.9
= 0.1
所以，P(第二次未擊中、其他三次都擊中)
= P(第一次擊中) X P(第二次未擊中) X P(第三次擊中) X P(第四次擊中)
= 0.9 X 0.1 X 0.9 X 0.9
= 0.0729
= 729 / 10 000
參考: Myself~~~
2007-07-21 8:31 pm
1.
P(both are " pass ")
= 96/100 x 97/100
= 582/625
2(1).
P(both places rain) = 0.2 x 0.3 = 0.06
2(2).
P(both places don't rain) = 0.8 x 0.7 = 0.56
2(3)
P(At least one place rains) = 1 - 0.06 - 0.56 = 0.38
3.P(Hitting the target in the remaining 3 times)
=0.9 x 0.9 x 0.9
=0.729
P(both are " pass ")
= 96/100 x 97/100
= 582/625
2(1).
P(both places rain) = 0.2 x 0.3 = 0.06
2(2).
P(both places don't rain) = 0.8 x 0.7 = 0.56
2(3)
P(At least one place rains) = 1 - 0.06 - 0.56 = 0.38
3.P(Hitting the target in the remaining 3 times)
=0.9 x 0.9 x 0.9
=0.729
收錄日期: 2021-04-13 00:50:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070721000051KK01418