小宇 asked in 科學數學 · 1 decade ago

研究所統計學問題part1 \”超級無敵霹靂急\”

1.A standard deck of cards is being used to play a game. There are four suits (hearts, diamonds, clubs, and spades ) , each having 13 cards (ace,2,3…..10,jack,queen,king)

making a total of 52 cards. This complete deck is thoroughly mixed, and you will receive the first two cards from the deck without replacement.

(1) what is the probability that both cards are queens?

(A)12/2652 (B)16/2652 (C)156/2652 (D)169/2652 (E)none of the above

(2) what is the probability that the first cards is 10 and the second card is a 5 or6?

(A)24/2652 (B)32/2652 (C)24/2704 (D) 32/2704 (D) 32/2704 (E)none of the above

(3) if we were sampling with replacement, what would be the answer in question(1)?

(A)12/2704 (B)16/2704 (C)156/2704 (D)169/2652 (E)none of the above

From 政大財管所

(2)設A、B為任意的兩事件,請證明P(A∩B)≧1-P(A¢)-P(B¢),其中A¢、B¢分別代表A、B的餘事件(complementary event)。

From 雲科大企研資管所

(4) 如果你和你的同學有三十人,請問至少有兩人是同月同日生的機率?

From 暨南經研所

(5) If P(B)=0.30 , P(A|B)=0.40 , and P(B|A)=0.50 , what dose P(A) epual?

(A) 0.10 (B)0.12 (C)0.30 (D)0.24 (E)can not be determined

From 交大科技管理所

Update:

希望會寫的人能夠將答案放上來,解幾題算幾題,盡量寫.

還有

研究所統計學問題part2 \”超級無敵霹靂急\”

2 Answers

Rating
  • ?
    Lv 7
    1 decade ago
    Favorite Answer

    1(1)4/52*3/51=12/2652=>A

    (2)4/52*8/51=32/2652=>B

    (3)4/52*4/52=16/2704

    2 P(A')=1-P(A),P(B')=1-P(B)

    P(A')+P(B')+P(A∩B)=P(A∩B)+1-P(A)+1-P(B)>=2-[P(A)+P(B)-P(A∩B)]>=2-1=1

    P(A∩B)≧1-P(A')-P(B')

    4 1-1*(1-1/365)(1-2/365)*...*(1-29/365)=70.6%

    5 P(A|B)=P(A∩B)/P(B),P(B|A)=P(A∩B)/P(A)

    0.4*0.3=P(A∩B)

    P(A∩B)=0.12

    So P(A)=0.12/0.5=12/50=0.24=>D

  • 1 decade ago

    (1)(A)

    (2) (B)

    (3) (B)

    (2 ) P(A聯集B)=P(A)+ P(B)- P(A∩B)<=1

    則P(A∩B) ≧P(A)+ P(B)-1=1-(1- P(A))-( 1- P(B))=1- P(A)- P(B)

    (4) 如果你和你的同學有三十人,請問至少有兩人是同月同日生的機率?

    1-P(三十人生日都不同)=1-P(365,30)/(365^30)=0.706….

    (5) (D)0.24

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