有两道数学关于概率论的题目不太会 题目是很简单的英文 我就不翻译了,求高人帮助!1. I form a 3-digit number at random from the digits 1,2,3,..., 9; determine (work your answers out completely) the probability that my nu

有两道数学关于概率论的题目不太会 题目是很简单的英文 我就不翻译了,求高人帮助!1. I form a 3-digit number at random from the digits 1,2,3,..., 9; determine (work your answers out completely) the probability that my nu
有两道数学关于概率论的题目不太会 题目是很简单的英文 我就不翻译了,求高人帮助!
1. I form a 3-digit number at random from the digits 1,2,3,..., 9; determine (work your answers out completely) the probability that my number has ...
(a) No repeated digits.
(b) All three digits the same.
(c) Exactly one repeated digit (i.e. two digits the same and one dierent).
Don't forget to work your answers out completely.
2. A bag contains 10 marbles: 1 red, 2 blue, 3 green, and 4 yellow. I reach into the bag and pick 5 marbles at random. Determine the probability that my pick has ...
(a) At least 2 yellow marbles.
(b) At least 2 yellow marbles and at least 1 green marble.
(c) All four colors represented.
(d) At least two colors represented.
(e) At most three colors represented.
Leave your answers in factorial notation.

有两道数学关于概率论的题目不太会 题目是很简单的英文 我就不翻译了,求高人帮助!1. I form a 3-digit number at random from the digits 1,2,3,..., 9; determine (work your answers out completely) the probability that my nu
1.
(a) No repeated digits. P(9,3)/9^3=9*8*7/9^3=56/81
(b) All three digits the same. C(9,1)/9^3=9/9^3=1/81
(c) Exactly one repeated digit. 1- 56/81-1/81=24/81=8/27

2.
(a) At least 2 yellow marbles.
1-C(6,5)/C(10,5)-C(4,1)*C(6,4)/C(10,5)=1-1/42-10/42=31/42
(b) At least 2 yellow marbles and at least 1 green marble.
1-C(7,5)/C(10,5)-C(6,5)/C(10,5)-C(4,1)*C(6,4)/C(10,5)
=1-7/84-1/42-10/42
=55/84
(c) All four colors represented.
C(1,1)*C(2,1)*C(3,1)*C(4,1)/C(10,5)
=1*2*3*4*(1*2*3*4*5)/(10*9*8*7*6)
=2/21
(d) At least two colors represented.
1-C(9,5)/C(10,5)-C(8,5)/C(10,5)-C(7,5)/C(10,5)-C(6,5)/C(10,5)
=1-1/2-2/9-1/12-1/42
=209/252
(e) At most three colors represented.
1-C(1,1)*C(2,1)*C(3,1)*C(4,1)/C(10,5)
=1-2/21
=19/21

1.事件的总可能数=9*9*9=729
(a)三个数字全不重复，P=(A上标3下标9)/729=56/81
(b)三个数字相同 ，P=9/729=1/81
(c)只有两个数字相同，P=(C上标1下标9)*（C上标1下标8）*3/729=8/27

2.事件的总可能数=C上标5下标10=252
(a)P=(C上标2下标4*C上标3下标6+C上标3下...

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1.事件的总可能数=9*9*9=729
(a)三个数字全不重复，P=(A上标3下标9)/729=56/81
(b)三个数字相同 ，P=9/729=1/81
(c)只有两个数字相同，P=(C上标1下标9)*（C上标1下标8）*3/729=8/27

2.事件的总可能数=C上标5下标10=252
(a)P=(C上标2下标4*C上标3下标6+C上标3下标4*C上标2下标6+C上标4下标4*C上标1下标6)/252
(b)P=(C上标2下标4*C上标1下标3*C上标2下标3+C上标3下标4*C上标1下标3*C上标1下标3+C上标4下标4*C上标1下标3+C上标2下标4*C上标2下标3*C上标1下标3+C上标3下标4*C上标2下标3+C上标2下标4*C上标3下标3)/252
(c)P=(3*4+2*C上标2下标3*4+C上标2下标4*2*3)/252
(d)P=1
(e)P=1-(3*4+2*C上标2下标3*4+C上标2下标4*2*3)/252

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