locmsimon asked in 科學及數學數學 · 9 years ago

數學題(Probability) 20分

8. At a school party, ten students sit randomly around a circular table that seats ten people. What is the probability that Janna and Roberto are not sitting opposite or beside each other?

11. Hockey Two sister, Zana and Kari,are among 15 girls trying out for forward positions on a hockey team, On the first day of practice, the coach ramdomly sets up five forward lines of three players each. If the coach differentiates between the postions of right wing, left wing, and center,

what is the probability that Zana and Kari are on the same line?

12. Bridge In the game of bridge, four players are dealt 13 cards each from a well-shuffled deck of 52 playing card.

b) What is the probability that each of the four players receives a complete suit in his or her hand?

d) What is the probability that the player who deals does not have a face card, or an ace, in his or her hand?

ans: 8=2/3 11=1/7 12b=4.47*10^28 12d=0.36%

主要:請解釋步驟點得出,or 可比意見


1 Answer

  • Favorite Answer

    8) suppose that Janna takes a position randomly, say position 1, then Roberto should not take position 2, 6 or 10 so that they won't sit opposite or beside each other, remaining a total of 6 possibilities.

    Hence the prob. is 6/9 = 2/3

    9) Suppose that Zana takes a line randomly, say line 1, then there are still 2 positions in line 1 out of a total of 14 positions.

    Thus if Kari takes any one of the positions in line 1, they will be on the same line, giving a prob. of 2/14 = 1/7.

    12b) When 52 cards are randomly arranged, no. of possible permutations = 52!

    So if we arrange in the manner:

    13 diamonds, 13 spades, 13 hearts, 13 clubs

    No. of possible permutations = (13!)4

    Then the suits can also be randomly arranged, giving a possible 4! = 24 permutations.

    So no. of possible permutations that each of the four players receives a complete suit in his or her hand is 24 x (13!)4

    So the required prob is 24 x (13!)4/52! = 4.47 x 10-28

    d) No. of combinatinos that contain no card or ace card is 36C13

    So the prob is 36C13/52C13 = 0.36%

    Source(s): 原創答案
Still have questions? Get your answers by asking now.