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!!Probability Problems!!?

Answer the question in detail please. ^v^

1) If the odds in favour of Ann beating Sue in a chess game are 5:4, what is the probability that Sue will win an upset victory in a best-of-five chess tournament?

2) Suppose the odds of the Toronto Maple Leafs winning the Stanley Cup are 1:5, while the odds of the Montreal Canadiens winning the Stanley Cup are 2:13. What are the odds in favour of either Tonronto or Montreal winning the Stanley Cup?

3) 4 friends, 2 girls and 2 boys, are playing contract bridge. Partners are randomly assigned for each game. What is the probability that the 2 girls will be partners for the first game?

4) Ann,Sue,Amy,Ken,and Kim are going to a party. What is the probability that 2 of the girls will arrive first?

5) A hockey team has 2 goalies,6 defenders,8 wingers,and 4 centres. If the team ramdomly select 4 players to attend a charity function, what is the likehood that no goalies or centres are selected?

1 Answer

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  • 1 decade ago
    Favorite Answer

    q1

    what's upset victory?

    i'll just make up the rule that upset victory in best of 5 is winning all last 5 from 9 games.

    so the winning sequence=AAAASSSSS; the only way it happen

    P=(5/9)^4(4/9)^5

    =4*20^4/9^9

    =640000/387420489

    =0.001652

    =0.1652%

    edit q1

    i just feel like changing make-up rules a bit. best of 5 needs you to win 3 to win the game. and this time sue get to win in many ways.

    P(sue wins the game)

    =P(in 5 sets)+P(in 4 sets)+P(in 3 sets)

    =5C3s^3*a^2 + 4C3s^3*a + 3C3s^3

    =[10(5/9)^2 + 4(5/9) + 1]*(4/9)^3

    =[250+180+81]*64/9^5

    =511*64/9^5

    =0.553845

    =55.3845%

    q2

    TML=1:5=1/6=5/30

    MC=2:13=2/15=4/30

    P=1/(1+5) + 2/(2+13)

    =1/6+2/15

    =(5+4)/30

    =9/30

    =3/10

    odds favouring either one

    =3:(10-3)

    =3:7

    q3

    how do you play contract bridge?

    P=2*2C2/4C2

    =2/6

    =1/3

    q4

    sorry, but i'm not good with names. Kim's a girl or boy? Kim's my surname, so i dont know. we'll assume ken's the only boy here.

    P=4*3*3*2*1/5*4*3*2*1

    =3/5

    q5

    likelihood?

    2 goalies,6 defenders,8 wingers,and 4 centres

    P(no goalies or centers)

    =(6+8)C4/(2+6+8+4)C4

    =14C4/20C4

    =11*12*13*14/17*18*19*20

    =11*13*7/17*19*15

    =1001/4845

    =0.2066

    =20.66%

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